模板积累
FAQ
- 绝大部分的RE错误来源于数组开的过小
- 注意整数除以整数的时候乘以1.0,除非要是用整除的含义(拓展欧几里得)
- 使用并查集的时候最后要扫一遍,使路径压缩。
- 时刻注意数字的范围，如果题目中出现了long long 一定要高度的重视，因为很有可能会被坑
- 开比较大的栈:#pragma comment(linker, “/STACK:102400000,102400000”)
快速读入挂
有时候并不一定就很快emmm
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| ll readll(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int readint(){ int x=0,f=1;char ch=getchar(); while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} while(isdigit(ch))x=x*10+ch-'0',ch=getchar(); return f*x; }
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c++高精度
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| #include<string> #include<iostream> #include<iosfwd> #include<cmath> #include<cstring> #include<stdlib.h> #include<stdio.h> #include<cstring> #define MAX_L 2005 using namespace std; class bign { public: int len, s[MAX_L]; bign(); bign(const char*); bign(int); bool sign; string toStr() const; friend istream& operator>>(istream &,bign &); friend ostream& operator<<(ostream &,bign &); bign operator=(const char*); bign operator=(int); bign operator=(const string); bool operator>(const bign &) const; bool operator>=(const bign &) const; bool operator<(const bign &) const; bool operator<=(const bign &) const; bool operator==(const bign &) const; bool operator!=(const bign &) const; bign operator+(const bign &) const; bign operator++(); bign operator++(int); bign operator+=(const bign&); bign operator-(const bign &) const; bign operator--(); bign operator--(int); bign operator-=(const bign&); bign operator*(const bign &)const; bign operator*(const int num)const; bign operator*=(const bign&); bign operator/(const bign&)const; bign operator/=(const bign&); bign operator%(const bign&)const; bign factorial()const; bign Sqrt()const; bign pow(const bign&)const; void clean(); ~bign(); }; #define max(a,b) a>b ? a : b #define min(a,b) a<b ? a : b bign::bign() { memset(s, 0, sizeof(s)); len = 1; sign = 1; } bign::bign(const char *num) { *this = num; } bign::bign(int num) { *this = num; } string bign::toStr() const { string res; res = ""; for (int i = 0; i < len; i++) res = (char)(s[i] + '0') + res; if (res == "") res = "0"; if (!sign&&res != "0") res = "-" + res; return res; } istream &operator>>(istream &in, bign &num) { string str; in>>str; num=str; return in; } ostream &operator<<(ostream &out, bign &num) { out<<num.toStr(); return out; } bign bign::operator=(const char *num) { memset(s, 0, sizeof(s)); char a[MAX_L] = ""; if (num[0] != '-') strcpy(a, num); else for (int i = 1; i < strlen(num); i++) a[i - 1] = num[i]; sign = !(num[0] == '-'); len = strlen(a); for (int i = 0; i < strlen(a); i++) s[i] = a[len - i - 1] - 48; return *this; } bign bign::operator=(int num) { char temp[MAX_L]; sprintf(temp, "%d", num); *this = temp; return *this; } bign bign::operator=(const string num) { const char *tmp; tmp = num.c_str(); *this = tmp; return *this; } bool bign::operator<(const bign &num) const { if (sign^num.sign) return num.sign; if (len != num.len) return len < num.len; for (int i = len - 1; i >= 0; i--) if (s[i] != num.s[i]) return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i])); return !sign; } bool bign::operator>(const bign&num)const { return num < *this; } bool bign::operator<=(const bign&num)const { return !(*this>num); } bool bign::operator>=(const bign&num)const { return !(*this<num); } bool bign::operator!=(const bign&num)const { return *this > num || *this < num; } bool bign::operator==(const bign&num)const { return !(num != *this); } bign bign::operator+(const bign &num) const { if (sign^num.sign) { bign tmp = sign ? num : *this; tmp.sign = 1; return sign ? *this - tmp : num - tmp; } bign result; result.len = 0; int temp = 0; for (int i = 0; temp || i < (max(len, num.len)); i++) { int t = s[i] + num.s[i] + temp; result.s[result.len++] = t % 10; temp = t / 10; } result.sign = sign; return result; } bign bign::operator++() { *this = *this + 1; return *this; } bign bign::operator++(int) { bign old = *this; ++(*this); return old; } bign bign::operator+=(const bign &num) { *this = *this + num; return *this; } bign bign::operator-(const bign &num) const { bign b=num,a=*this; if (!num.sign && !sign) { b.sign=1; a.sign=1; return b-a; } if (!b.sign) { b.sign=1; return a+b; } if (!a.sign) { a.sign=1; b=bign(0)-(a+b); return b; } if (a<b) { bign c=(b-a); c.sign=false; return c; } bign result; result.len = 0; for (int i = 0, g = 0; i < a.len; i++) { int x = a.s[i] - g; if (i < b.len) x -= b.s[i]; if (x >= 0) g = 0; else { g = 1; x += 10; } result.s[result.len++] = x; } result.clean(); return result; } bign bign::operator * (const bign &num)const { bign result; result.len = len + num.len; for (int i = 0; i < len; i++) for (int j = 0; j < num.len; j++) result.s[i + j] += s[i] * num.s[j]; for (int i = 0; i < result.len; i++) { result.s[i + 1] += result.s[i] / 10; result.s[i] %= 10; } result.clean(); result.sign = !(sign^num.sign); return result; } bign bign::operator*(const int num)const { bign x = num; bign z = *this; return x*z; } bign bign::operator*=(const bign&num) { *this = *this * num; return *this; } bign bign::operator /(const bign&num)const { bign ans; ans.len = len - num.len + 1; if (ans.len < 0) { ans.len = 1; return ans; } bign divisor = *this, divid = num; divisor.sign = divid.sign = 1; int k = ans.len - 1; int j = len - 1; while (k >= 0) { while (divisor.s[j] == 0) j--; if (k > j) k = j; char z[MAX_L]; memset(z, 0, sizeof(z)); for (int i = j; i >= k; i--) z[j - i] = divisor.s[i] + '0'; bign dividend = z; if (dividend < divid) { k--; continue; } int key = 0; while (divid*key <= dividend) key++; key--; ans.s[k] = key; bign temp = divid*key; for (int i = 0; i < k; i++) temp = temp * 10; divisor = divisor - temp; k--; } ans.clean(); ans.sign = !(sign^num.sign); return ans; } bign bign::operator/=(const bign&num) { *this = *this / num; return *this; } bign bign::operator%(const bign& num)const { bign a = *this, b = num; a.sign = b.sign = 1; bign result, temp = a / b*b; result = a - temp; result.sign = sign; return result; } bign bign::pow(const bign& num)const { bign result = 1; for (bign i = 0; i < num; i++) result = result*(*this); return result; } bign bign::factorial()const { bign result = 1; for (bign i = 1; i <= *this; i++) result *= i; return result; } void bign::clean() { if (len == 0) len++; while (len > 1 && s[len - 1] == '\0') len--; } bign bign::Sqrt()const { if(*this<0)return -1; if(*this<=1)return *this; bign l=0,r=*this,mid; while(r-l>1) { mid=(l+r)/2; if(mid*mid>*this) r=mid; else l=mid; } return l; } bign::~bign() { } bign num[105]; int main() { num[1] = 1, num[2] = 5; int n; cin>>n; for(int i=3; i<=n; i++){ num[i] = num[i-1]*3-num[i-2]+2; } cout << num[n] << endl; return 0; }
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int128的读入
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| #include <bits/stdc++.h> using namespace std; template <class T> inline bool scan_d(T & ret) { char c; int sgn; if(c = getchar(), c == EOF)return false; while(c != '-' && (c < '0' || c > '9'))c = getchar(); sgn = (c == '-') ? -1 : 1; ret = (c == '-') ? 0 : (c - '0'); while(c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0'); ret *= sgn; return true; } #ifdef Cpp11 template <class T, class ... Args> inline bool scan_d(T & ret, Args & ... args) { scan_d(ret); scan_d(args...); } #define cin.tie(0); cin.tie(nullptr); #define cout.tie(0); cout.tie(nullptr); #endif inline bool scan_ch(char &ch) { if(ch = getchar(), ch == EOF)return false; while(ch == ' ' || ch == '\n')ch = getchar(); return true; } template <class T> inline void out_number(T x) { if(x < 0) { putchar('-'); out_number(- x); return ; } if(x > 9)out_number(x / 10); putchar(x % 10 + '0'); } int main() { __int128 a, b; scan_d(a), scan_d(b); out_number(a + b); puts(""); return 0; }
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python 读入
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| import math if __name__ == '__main__': try: while True: a, b, c = map(int, input().split()) print(pow(a, int(pow(b, c, int(1e9 + 6))), int(1e9 + 7))) except EOFError: pass
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读取带有空格的字符串
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| //method 1 string str; getline(cin, str); //method 2 char str[maxn]; gets(str);
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adptive Simpson’s rule
为什么求sin(x)/x就会挂掉啊不知道MATLAB能不能求出来这种形式的积分，明天试试。
还有三重积分球体积怎么用数值分析的方法来求？
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| #include <bits/stdc++.h> using namespace std; const double PI = acos(-1.0); double F(double x){ return sin(x); } double simpson(double a, double b){ double c = a+(b-a)/2; return (F(a)+4*F(c)+F(b))*(b-a)/6; } double asr(double a, double b, double eps, double A){ double c = a+(b-a)/2; double L = simpson(a, c), R = simpson(c, b); if(fabs(L+R-A) <= 15*eps) return L+R+(L+R-A)/15.0; return asr(a, c, eps/2, L)+asr(c, b, eps/2, R); } double asr(double a, double b, double eps){ return asr(a, b, eps, simpson(a, b)); } int main() { double s = asr(0, PI, 1e-5); cout<<s<<endl; return 0; }
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二分图匹配
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| #include <bits/stdc++.h> using namespace std; const int maxn = 505; vector<int> G[maxn*2]; int match[maxn*2]; int used[maxn*2]; int n, m, l; void add_edge(int u, int v) { G[u].push_back(v); G[v].push_back(u); } bool dfs(int v){ used[v] = true; for(int i=0; i<G[v].size(); i++){ int u = G[v][i], w = match[u]; if(w == -1 || !used[w]&&dfs(w)){ match[u] = v; match[v] = u; return true; } } return false; } void solve(){ int ans = 0; memset(match, -1, sizeof(match)); for(int v=0; v<=n; v++){ if(match[v] == -1){ memset(used, 0, sizeof(used)); if(dfs(v)) ans++; } } for(int v=500; v<=m+500; v++){ if(match[v] == -1){ memset(used, 0, sizeof(used)); if(dfs(v)) ans++; } } printf("%d\n", ans); } int main() { while(~scanf("%d%d", &n, &m)){ for(int i=0 ;i<maxn*2; i++) G[i].clear(); scanf("%d", &l); for(int i=1; i<=l; i++){ int u, v; scanf("%d%d", &u, &v); add_edge(u, v+500); } solve(); } return 0; }
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Dinic算法求最大流
接口
n:节点的个数
m:边的个数
s:起始点(源点)
t:终点(汇点)
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| #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; const int maxn = 205; const int INF = 1e9+10; struct Edge{ int to, cap, rev; }; vector<Edge> G[maxn]; int cnt[maxn]; int level[maxn]; int s, t; int n, m; void add_edge(int u, int v, int cap){ G[u].push_back(Edge{v, cap, G[v].size()}); G[v].push_back(Edge{u, 0, G[u].size()-1}); } void bfs(int s){ memset(level, -1, sizeof(level)); queue<int> q; q.push(s); level[s] = 0; while(!q.empty()){ int v = q.front(); q.pop(); for(int i=0; i<G[v].size(); i++){ Edge& e = G[v][i]; if(e.cap>0&&level[e.to]<0){ level[e.to] = level[v]+1; q.push(e.to); } } } } int dfs(int v, int t, int flow){ if(v == t) return flow; for(int& i=cnt[v]; i<G[v].size(); i++){ Edge& e = G[v][i]; if(e.cap>0&&level[v]<level[e.to]){ int d = dfs(e.to, t, min(flow, e.cap)); if(d>0){ e.cap -= d; G[e.to][e.rev].cap += d; return d; } } } return 0; } int max_flow(int s, int t){ int flow = 0; for(;;){ bfs(s); if(level[t]<0) return flow; int f; memset(cnt, 0, sizeof(cnt)); while((f = dfs(s, t, INF))>0){ flow+=f; } } } int main() { while(~scanf("%d%d", &n, &m)){ for(int i=0; i<=n; i++) G[i].clear(); int u, v, cap; for(int i=0; i<m; i++){ scanf("%d %d %d", &u, &v, &cap); add_edge(u, v, cap); } s = 1; t = n; printf("%d\n", max_flow(s, t)); } return 0; }
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最小费用流最新模板
费用是float类型的。
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| #include <cstdio> #include <algorithm> #include <cstring> #include <vector> #include <queue> #include <cmath> #include <iostream> #define double float using namespace std; const int maxn = 200+10; namespace mincostflow { const int INF=0x3f3f3f3f; struct node { int to; int cap; double cost; int rev; node(int t=0,int c=0,double _c=0,int n=0): to(t),cap(c),cost(_c),rev(n) {}; }; vector<node> edge[maxn]; void addedge(int from,int to,int cap,double cost) { edge[from].push_back(node(to,cap,cost,edge[to].size())); edge[to].push_back(node(from,0,-cost,edge[from].size()-1)); } double dis[maxn]; bool mark[maxn]; void spfa(int s,int t,int n) { for(int i=0; i<=n+2; i++) dis[i] = double (INF); memset(mark+1,0,n*sizeof(bool)); static int Q[maxn],ST,ED; dis[s]=0; ST=ED=0; Q[ED++]=s; while (ST!=ED) { int v=Q[ST]; mark[v]=0; if ((++ST)==maxn) ST=0; for (node &e:edge[v]) { if (e.cap>0&&dis[e.to]>dis[v]+e.cost) { dis[e.to]=dis[v]+e.cost; if (!mark[e.to]) { if (ST==ED||dis[Q[ST]]<=dis[e.to]) { Q[ED]=e.to,mark[e.to]=1; if ((++ED)==maxn) ED=0; } else { if ((--ST)<0) ST+=maxn; Q[ST]=e.to,mark[e.to]=1; } } } } } } int cur[maxn]; int dfs(int x,int t,int flow) { if (x==t||!flow) return flow; int ret=0; mark[x]=1; for (int &i=cur[x];i<(int)edge[x].size();i++) { node &e=edge[x][i]; if (!mark[e.to]&&e.cap) { if (dis[x]+e.cost==dis[e.to]) { int f=dfs(e.to,t,min(flow,e.cap)); e.cap-=f; edge[e.to][e.rev].cap+=f; ret+=f; flow-=f; if (flow==0) break; } } } mark[x]=0; return ret; } pair<int,double> min_costflow(int s,int t,int n) { int ret=0; double ans=0; int flow = INF; while (flow) { spfa(s,t,n); if (dis[t]==double(INF)) break; memset(cur+1,0,n*sizeof(int)); double len=dis[t]; int f; while ((f=dfs(s,t,flow))>0) ret+=f,ans+=len*f,flow-=f; } return make_pair(ret,ans); } void init(int n) { int i; for (int i = 1; i <= n; i++) edge[i].clear(); } } int n, m; using namespace mincostflow; int ss, tt; double e = 2.718281828459045; int main(int argc, char const *argv[]) { int _; scanf("%d", &_); while(_--){ scanf("%d %d", &n, &m); init(n+2); int x, y, z; double s; ss=1, tt = n+2; for(int i=1; i<=n; i++) { scanf("%d%d", &x, &y); if(x) addedge(ss, i + 1, x, 0); if(y) addedge(i + 1, tt, y, 0); } for (int i = 1; i <= m; i++) { scanf("%d %d %d %f", &x, &y, &z, &s); if(z)addedge(x + 1, y + 1, z-1, -log(1.0 - s)), addedge(x + 1, y + 1, 1, 0.0); } pair<int,double > ans = min_costflow(ss, tt, n+2); printf("%.2f\n", 1.0-pow(e, -ans.second)); } return 0; }
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链式前向星来存图
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| #include <bits/stdc++.h> using namespace std; const int maxn = 4e5+100; struct node{ int to, next, weight; }edge[maxn]; int cnt; int head[maxn]; int n; void add(int u, int v, int weight) { edge[cnt].to = v; edge[cnt].weight = weight; edge[cnt].next = head[u]; head[u] = cnt++; } int main() { while(cin>>n){ cnt=0, memset(head, -1, sizeof(head)); for(int i=0; i<n-1; i++){ int u, v, weight; cin>>u>>v>>weight; add(u, v, weight); add(v, u, weight); } for(int i=1; i<=n; i++){ for(int j=head[i]; j!=-1; j = edge[j].next){ cout<<edge[j].to<<" "; } cout<<endl; } } return 0; }
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快速幂
快速幂
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| #include <cstdio> using namespace std; typedef long long ll; ll a, b, c; ll pow_mod(ll a, ll b, ll mod){ ll ans = 1; while(b){ if(b&1){ ans = ans*a%mod; } a = a*a%mod; b>>=1; } return ans; } int main(){ while(~scanf("%I64d%I64d%I64d", &a, &b, &c)){ printf("%I64d\n", pow_mod(a, b, c)); } return 0; }
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矩阵快速幂
斐波那契数列
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| #include <cstdio> #include <cstring> using namespace std; typedef long long ll; const int maxn = 2; const int mod = 7; int A, B, n; struct Mat{ int mat[maxn][maxn]; Mat(){ memset(mat, 0, sizeof(mat)); } }; Mat mul(Mat a, Mat b, int n){ Mat ans; for(int i=0; i<n; i++){ for(int k=0; k<n; k++){ for(int j=0; j<n; j++){ ans.mat[i][j] = (ans.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod; } } } return ans; } Mat pow_mat(Mat a, int b, int n){ Mat ans; for(int i=0; i<n; i++)ans.mat[i][i] = 1; while(b){ if(b&1){ ans = mul(a, ans, n); } a = mul(a, a, n); b>>=1; } return ans; } int main(){ while(~scanf("%d%d%d", &A, &B, &n)&&A+B+n!=0){ Mat a; a.mat[0][0] = A, a.mat[0][1] = B, a.mat[1][0] = 1, a.mat[1][1] = 0; if(n <= 2){ printf("%d\n", 1); } else{ Mat ans = pow_mat(a, n-2, 2); int temp = (ans.mat[0][0]+ans.mat[0][1])%mod; printf("%d\n", temp); } } return 0; }
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求逆元
拓展欧几里德求解逆元
用拓展欧几里德来求解逆元的时候，需要\(gcd(b, mod)=1\)。
拓展欧几里德的模板题
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| #include <bits/stdc++.h> using namespace std; typedef long long ll; const int mod = 9973; void extgcd(ll a,ll b,ll& d,ll& x,ll& y){ if(!b){ d=a; x=1; y=0;} else{ extgcd(b,a%b,d,y,x); y-=x*(a/b); } } ll inverse(ll a,ll n){ ll d,x,y; extgcd(a,n,d,x,y); return d==1?(x+n)%n:-1; } int T; int n, b; int main() { scanf("%d", &T); for(int i=0; i<T; i++){ scanf("%d%d", &n, &b); ll inver = inverse(b, mod); int ans = inver*n%mod; printf("%d\n", ans); } return 0; }
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费马小定理求解逆元
若被摸的数p为素数的时候，有拓展费马小定理证明：
\(a^{p-1}\equiv 1mod p\)
那么a的逆元就为\(a^{p-2}\).
快速幂搞一下就行了
一般的求解。
一般求解方法：
}{b} "\frac{a}{b}\%p=\frac{a\%(p*b)}{b}")
但是p*b的值往往是比较的大的，有时候会溢出就不知道怎么弄了。(看后面的有关的一道数论的题目)
c++中一些优秀的函数
strstr
可以部分的取代kmp算法，但是不能统计子串的个数。
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; char pattern[maxn]; char text[maxn<<4]; int main() { char *ptr; while(~scanf("%s%s", pattern, text)){ ptr = strstr(text, pattern); int len = strlen(pattern); if(ptr != NULL){ char ans[maxn]; for(int i=0; i<len; i++){ ans[i] = ptr[i]; } ans[len] = '\0'; printf("%s\n", ans); } else{ printf("-1\n"); } } return 0; }
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unique
使用unique之前要先对数组进行相应的排序，然后求得不同数字的个数。
unique不改变原来数组
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| #include <bits/stdc++.h> using namespace std; int main(){ int num[20]; for(int i=0; i<10; i++){ num[i] = i+1; } for(int i=10; i<20; i++){ num[i] = i-9; } sort(num, num+20); int sz = unique(num, num+20)-num; cout<<sz<<endl; return 0; }
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lower_bound()
返回第一个大于等于这个值的下标（下标是从0开始的）
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| int num[10] = {1, 2, 3, 4, 5, 5, 6, 7, 8, 9}; printf("%d\n", lower_bound(num, num+10, 5)-num);
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upper_bound()
返回第一个大于这个值的下标
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| int num[10] = {1, 2, 3, 4, 5, 5, 6, 7, 8, 9}; printf("%d\n", upper_bound(num, num+10, 5)-num);
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make_heap&&distance
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| #include <bits/stdc++.h> using namespace std; const int maxn = 55; int n; int num[maxn]; vector<int> heap; int main() { scanf("%d", &n); for(int i=0; i<n; i++){ scanf("%d", &num[i]); heap.push_back(num[i]); } make_heap(heap.begin(), heap.end(), greater<int>() ); for(vector<int>::iterator it = heap.begin(); it!=heap.end(); it++){ printf("%d ", *it); } printf("\n"); return 0; } #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; vector<int> num; int dis(int number){ return distance(num.begin(), find(num.begin(), num.end(), number)); } int main(){ int n; for(int i=1; i<=10; i++){ num.push_back(i); } cout<<dis(5)<<endl; }
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bitset
bitset的各种操作
高中OIer真的是惹不起，太强了。初中就已经到达了我现在的水平，真的是好惭愧啊。
bitset的长度在刚开始的时候就已经定义好了
priority_queue
符号重载后面加const的原因
简单来说在这个函数里面不能修改任何的数据，否则就会报错，降低错误率。
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| #include <bits/stdc++.h> using namespace std; struct Node{ string name; int top; friend bool operator < (Node a, Node b){ return a.top>b.top; } }; priority_queue<Node> pq; struct cmp{ bool operator () (Node a, Node b){ if(a.top!=b.top) return a.top>b.top; } }; priority_queue<Node, vector<Node>, cmp> pq2; int main(){ pq2.push(Node{"tian", 2}); pq2.push(Node{"bao", 1}); pq2.push(Node{"lin", 3}); Node temp = pq2.top(); cout<<temp.name<<" "<<temp.top<<endl; return 0; }
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树状数组
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; int lowbit(int x){ return x&-x; } int num[maxn]; int c[maxn]; int n; void update(int x, int val){ while(x<=n){ c[x] += val; x += lowbit(x); } } int sum(int x){ int ans = 0; while(x>0){ ans += c[x]; x -= lowbit(x); } return ans; } int main(){ while(~scanf("%d", &n)){ for(int i=1; i<=n; i++){ scanf("%d", &num[i]); } for(int i=1; i<=n; i++){ update(i, num[i]); } for(int i=1; i<=n; i++){ printf("%d: %d\n", i, sum(i)); } } return 0; }
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分治求逆序对的个数
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| ll merge_count(vector<int> &a){ int n = a.size(); if(n<=1) return 0; ll cnt = 0; vector<int> b(a.begin(), a.begin()+n/2); vector<int> c(a.begin()+n/2, a.end()); cnt += merge_count(b); cnt += merge_count(c); int ai = 0, bi = 0, ci=0; while(ai<n){ if(bi<b.size()&&(ci == c.size()||b[bi]<=c[ci])){ a[ai++] = b[bi++]; } else{ cnt += n/2-bi; a[ai++] = c[ci++]; } } return cnt; }
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KMP算法
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; int nexT[maxn]; char text[maxn]; char pattern[maxn]; int len1, len2; void get_next(){ nexT[0] = -1; int j = -1; for(int i=1; i<len2; i++){ while(j!=-1&&pattern[i]!=pattern[j+1]){ j = nexT[j]; } if(pattern[i] == pattern[j+1]){ j++; } nexT[i] = j; } } int KMP(){ get_next(); int j=-1; int ans = 0; for(int i=0; i<len1; i++){ while(j!=-1&&text[i]!=pattern[j+1]){ j = nexT[j]; } if(text[i] == pattern[j+1]){ j++; } if(j == len2-1){ ans++; j = nexT[j]; } } return ans; } int main() { while(~scanf("%s%s", text, pattern)){ len1 = strlen(text); len2 = strlen(pattern); printf("%d\n", KMP()); } return 0; }
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二分图的染色问题
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; int color[maxn]; vector<int> G[maxn]; int n, m; bool bipartite(int u){ for(int i=0; i<G[u].size(); i++){ int v = G[u][i]; if(color[v]&&color[v] == color[u]) return false; if(!color[v]){ color[v] = 3 - color[u]; if(!bipartite(v)) return false; } } return true; } int main() { while(~scanf("%d%d", &n, &m)){ for(int i=0; i<n; i++) G[i].clear(); memset(color, 0, sizeof(color)); int u, v; for(int i=0; i<m; i++){ scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } if(bipartite(0)){ printf("YES\n"); } else{ printf("NO\n"); } } return 0; }
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割顶
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; int n, m; vector<int> G[maxn]; int dfs_clock; int pre[maxn], low[maxn]; bool iscut[maxn]; int dfs(int u, int fa){ int lowu = pre[u] = ++dfs_clock; int child = 0; for(int i=0; i<G[u].size(); i++){ int v = G[u][i]; if(!pre[v]){ child++; int lowv = dfs(v, u); lowu = min(lowv, lowu); if(lowv>=pre[u]){ iscut[u] = true; } } else if(pre[u]>pre[v]&&fa!=v){ lowu = min(lowu, pre[v]); } } if(fa<0 &&child == 1) iscut[u] = false; low[u] = lowu; return lowu; } int main() { while(~scanf("%d%d", &n, &m)){ for(int i=0; i<n; i++) G[i].clear(); memset(iscut, 0, sizeof(iscut)); memset(pre, 0, sizeof(pre)); int u, v; for(int i=0; i<m; i++){ scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } for(int i=0; i<n; i++){ if(!pre[i]) dfs(i, -1); } for(int i=0; i<n; i++){ printf("%d ", iscut[i]); } printf("\n"); } return 0; }
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树的三序
还差一个层序遍历
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; int pre[maxn]; int in[maxn]; int post[maxn]; int n; struct Node{ int data; Node* lchild; Node* rchild; Node(){ this->data = -1; this->lchild = this->rchild = NULL; } }; Node* build_post(int prel, int prer, int inl, int inr){ if(prer<prel) return NULL; Node* root = new Node; root->data = pre[prel]; int temp = pre[prel]; int k; for(k=inl; k<=inr; k++){ if(temp == in[k]) break; } int numleft = k-inl; root->lchild = build_post(prel+1, prel+numleft, inl, k-1); root->rchild = build_post(prel+numleft+1, prer, k+1, inr); return root; } Node* build_pre(int postl, int postr, int inl, int inr){ if(postr<postl) return NULL; Node* root = new Node; root->data = post[postr]; int temp = pre[postr]; int k; for(k=inl; k<=inr; k++){ if(temp == in[k]) break; } int numleft = k-inl; root->lchild = build_pre(postl, postl+numleft-1, inl, k-1); root->rchild = build_pre(postl+numleft, postr-1, k+1, inr); return root; } void post_order(Node* root){ if(root == NULL) return; post_order(root->lchild); post_order(root->rchild); printf("%d", root->data); } void pre_order(Node* root){ if(root == NULL) return; printf("%d", root->data); pre_order(root->lchild); pre_order(root->rchild); } void layer_order(Node *root){ queue<Node*> q; q.push(root); while(!q.empty()){ Node *temp = q.front(); q.pop(); printf("%d ", temp->data); if(temp->lchild != NULL) q.push(temp->lchild); if(temp->rchild != NULL) q.push(temp->rchild); } } int main() { int type; printf("请输入个数n以及所得到的序列:\n1.先序和中序\n2.中序和后序\n"); while(~scanf("%d", &type)){ printf("请输入数组的大小\n"); scanf("%d", &n); if(type == 1){ for(int i=0; i<n; i++) scanf("%d", &pre[i]); for(int i=0; i<n; i++) scanf("%d", &in[i]); Node* root = build_post(0, n-1, 0, n-1); post_order(root); printf("\n"); } else if(type == 2){ for(int i=0; i<n; i++) scanf("%d", &post[i]); for(int i=0; i<n; i++) scanf("%d", &in[i]); Node* root = build_pre(0, n-1, 0, n-1); printf("444\n"); pre_order(root); printf("\n"); } else{ printf("你是不是想捣乱！\n"); break; } } return 0; }
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逆元
要求\(a\lt n\), 并且gcd(a, n) == 1
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| typedef long long ll; ll ex_gcd(ll a, ll b, ll &x, ll &y){ if(a==0 && b==0) return -1; if(b == 0){ x=1, y=0; return a; } ll d = ex_gcd(b, a%b, y, x); y -= a/b*x; return d; } ll inv(ll a, ll n){ ll x, y; ll d = ex_gcd(a, n, x, y); if(d == 1) return (x%n+n)%n; else return -1; }
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求组合数
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| typedef long long ll; const int maxn = 67; ll res[maxn][maxn]; ll Comb(ll n, ll m){ if(m == 0|| n == m) return 1; if(res[n][m]!=0) return res[n][m]; return res[n][m] = C(n-1, m-1)+C(n-1, m); }
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Lucas定理
p要求为素数，p<=1e5, n, m<=1e18
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| int n, m, p; typedef long long ll; ll ex_gcd(ll a, ll b, ll &x, ll &y){ if(a == 0&&b == 0) return -1; if(b ==0){ x=1, y=0; return a; } ll d=ecx_gcd(b, a%b, y, x); y -= a/b*x; return d; } ll inv(ll a, ll n){ ll x, y; ll d = ex_gcd(a, n, x, y); if(d!=1) return -1; else return (x%n+n)%n; } int C(int n, int m){ int ans = 1; for(int i=1; i<=m; i++){ ans = ans*(n-m+i)%p; ans = ans*inv(i, p)%p; } return ans; } int lucas(int n, int m){ if(m == 0) return 1; return C(n%p, m%p)*lucas(n/p, m/p)%p; }
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p不为质数的组合数求解
先将分子分母进行相应的质因子分解，然后消去分母，最后进行快速幂。
4C默写
搞完就删
堆排序
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; int heap[maxn]; int n; void downAdjust(int low, int high){ int i=low, j=2*i; while(j<=high){ if(j+1<=high&&heap[j+1]>heap[j]){ j++; } if(heap[j]>heap[i]){ swap(heap[i], heap[j]); i = j; j = i*2; } else break; } } void creat_heap(){ for(int i=n/2; i>=1; i--){ downAdjust(i, n); } } void deletetop(){ heap[1] = heap[n--]; downAdjust(1, n); } void upAdjust(int low, int high){ int i=high, j=i/2; while(j>=low){ if(heap[j]<heap[i]){ swap(heap[i], heap[j]); i = j; j = i/2; } else break; } } void insert_heap(int x){ heap[++n] = x; upAdjust(1, n); } void heapsort(){ creat_heap(); for(int i=n; i>=1; i--){ swap(heap[i], heap[1]); downAdjust(1, i-1); } } int main() { scanf("%d", &n); for(int i=1; i<=n; i++){ scanf("%d", &heap[i]); } creat_heap(); for(int i=1; i<=n; i++){ printf("%d ", heap[i]); } heapsort(); for(int i=1; i<=n; i++){ printf("%d ", heap[i]); } return 0; }
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树的层序遍历
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; struct Node{ int data; Node *lchild, *rchild; Node(){ this->data = -1; this->lchild = this->rchild = NULL; } }; int pre[maxn]; int in[maxn]; Node* build(int prel, int prer, int inl, int inr){ if(prel>prer) return NULL; Node *root = new Node; root->data = pre[prel]; int k; for(k=inl; k<=inr; k++){ if(pre[prel] == in[k]) break; } int num = k-inl; root->lchild = build(prel+1, prel+num, inl, k-1); root->rchild = build(prel+num+1, prer, k+1, inr); return root; } void post_order(Node *root){ if(root == NULL) return; post_order(root->lchild); post_order(root->rchild); printf("%d ", root->data); } void layer_order(Node *root){ queue<Node*> q; q.push(root); while(!q.empty()){ Node *temp = q.front(); q.pop(); printf("%d ", temp->data); if(temp->lchild != NULL) q.push(temp->lchild); if(temp->rchild != NULL) q.push(temp->rchild); } } int n; int main(){ scanf("%d", &n); for(int i=0; i<n; i++) scanf("%d", &pre[i]); for(int i=0; i<n; i++) scanf("%d", &in[i]); Node *root = build(0, n-1, 0, n-1); post_order(root); printf("\n"); layer_order(root); return 0; }
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最长回文串
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e3+10; string str; int dp[maxn][maxn]; int main() { getline(cin, str); memset(dp, 0, sizeof(dp)); int len = str.length(); int ans = 1; for(int i=0; i<len; i++){ dp[i][i] = 1; if(i+1<len){ if(str[i] == str[i+1]){ dp[i][i+1]=1; ans = 2; } } } for(int l=3; l<=len; l++){ for(int i=0; i+l-1<len; i++){ int j = i+l-1; if(str[i] == str[j] &&dp[i+1][j-1]){ dp[i][j] = 1; ans = l; } } } printf("%d\n", ans); }
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scc
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| #include <iostream> #include <cstring> #include <cstdio> #include <vector> using namespace std; const int maxn = 1e3+10; vector<int> G[maxn]; vector<int> rG[maxn]; bool vis[maxn]; int index[maxn]; vector<int> vs; void add_edge(int u, int v){ G[u].push_back(v); rG[v].push_back(u); } int n, m; void dfs(int u){ vis[u] = true; for(int i=0; i<G[u].size(); i++){ if(!vis[G[u][i]]) dfs(G[u][i]); } vs.push_back(u); } void rdfs(int v, int k){ vis[v] = true; index[v] = k; for(int i=0; i<rG[v].size(); i++){ if(!vis[rG[v][i]])rdfs(rG[v][i], k); } } int scc(){ memset(vis, 0, sizeof(vis)); vs.clear(); for(int i=0; i<n; i++){ if(!vis[i]) dfs(i); } memset(vis, 0, sizeof(vis)); int k = 0; for(int i=vs.size()-1; i>=0; i--){ if(!vis[vs[i]]){ rdfs(vs[i], k++); } } printf("%d\n", k); return k; } int main() { while(~scanf("%d%d", &n, &m)){ for(int i=0; i<n; i++){ G[i].clear(); rG[i].clear(); } int u, v; for(int i=0; i<m; i++){ scanf("%d%d", &u, &v); add_edge(u-1, v-1); } scc(); for(int i=0; i<n; i++){ printf("%d: %d\n", i, index[i]); } } return 0; }
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最长上升子序列
感觉运用了贪心的细想，不断的更新最长的子序列。
用二分查找的方法找到一个位置，使得\(num\gt b[i-1] 并且num\lt b[i],并用num代替b[i]\)
注意二分的总结
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| #include <bits/stdc++.h> using namespace std; const int maxn = 1e5+10; int a[maxn], b[maxn]; int n; int len; int bi(int num){ int L=1, R=len; while(L<=R){ int M = (L+R)/2; if(num>=b[M]) L = M-1; else R = M-1; } printf("len: %d\n", L); return L; } int main() { while(~scanf("%d", &n)){ for(int i=1; i<=n; i++) scanf("%d", &a[i]); len=1; b[1] = a[1]; for(int i=2; i<=n; i++){ if(a[i]>=b[len]){ b[++len] = a[i]; } else{ int pos = bi(a[i]); b[pos] = a[i]; } } printf("%d\n", len); for(int i=1; i<=len; i++){ printf("%d ", b[i]); } printf("\n"); } return 0; }
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