hdu 4902 nice boats

hdu 4902

HDU 4902 Nice boat

题目翻译

我们需要支持以下操作:

  • 1 L R C:将 $\forall i \in [L, R], \ A_i$ 替换成 $C$;
  • 2 L R C:将 $\forall i \in [L, R], \ A_i$ 替换成 $\min {A_i, \gcd(A_i, C)}$。

题目分析

开始感觉和 HDU 5306 很像,但是仔细想了想又不是那么回事…

题解

懒惰更新。如果这个区间里面的所有数都比 $C$ 小,就不用执行操作2了。所以我们需要维护区间最大值标记,如果最大值都比 $C$ 小,我们就停止更新。

输出的时候有一个小技巧:如果这个区间有标记,直接输出标记就行。

AC Code

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//#define NOSTDCPP
//#define Cpp11
//#define Linux_System
#ifndef NOSTDCPP
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstring>
#include <cstdio>
#include <deque>
#include <exception>
#include <functional>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#endif
# ifdef Linux_System
# define getchar getchar_unlocked
# define putchar putchar_unlocked
# endif
# define RESET(_) memset(_, 0, sizeof(_))
# define RESET_(_, val) memset(_, val, sizeof(_))
# define fi first
# define se second
# define pb push_back
# define midf(x, y) ((x + y) >> 1)
# define DXA(_) ((_ << 1))
# define DXB(_) ((_ << 1) | 1)
# define next __Chtholly__
# define x1 __Mercury__
# define y1 __bbtl04__
# define index __ikooo__
using namespace std;
typedef long long ll;
typedef vector <int> vi;
typedef set <int> si;
typedef pair <int, int> pii;
typedef long double ld;
const int MOD = 1e9 + 7;
const int maxn = 200009;
const int maxm = 300009;
const double pi = acos(-1.0);
const double eps = 1e-6;
ll myrand(ll mod){return ((ll)rand() << 32 ^ (ll)rand() << 16 ^ rand()) % mod;}
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;
}
inline void out_number(ll x)
{
if(x < 0)
{
putchar('-');
out_number(- x);
return ;
}
if(x > 9)out_number(x / 10);
putchar(x % 10 + '0');
}
struct node
{
int ma, tag;
//ll sum;
}tree[maxn << 2];
void pushup(int p, int l, int r)
{
if(l < r)
{
tree[p].ma = max(tree[DXA(p)].ma, tree[DXB(p)].ma);
}
}
void pushdown(int p, int l, int r)
{
if(~ tree[p].tag)
{
tree[DXA(p)].ma = tree[p].ma;
tree[DXA(p)].tag = tree[p].tag;
tree[DXB(p)].ma = tree[p].ma;
tree[DXB(p)].tag = tree[p].tag;
tree[p].tag = -1;
}
}
void pre(int l, int r, int p)
{
tree[p].ma = -1;
tree[p].tag = -1;
if(l == r)
{
//scan_d(tree[p].data);
scanf("%d", &tree[p].ma);
return ;
}
int mid = midf(l, r);
pre(l, mid, DXA(p));
pre(mid + 1, r, DXB(p));
pushup(p, l, r);
}
int l, r, n, m;
int x;
void update_1(int nl, int nr, int p)
{
if(l <= nl && nr <= r)
{
tree[p].tag = x;
tree[p].ma = x;
return ;
}
pushdown(p, nl, nr);
int mid = midf(nl, nr);
if(l <= mid) update_1(nl, mid, DXA(p));
if(mid < r) update_1(mid + 1, nr, DXB(p));
pushup(p, nl, nr);
}
void update_2(int nl, int nr, int p)
{
if(l <= nl && nr <= r)
{
if(tree[p].ma <= x) return ;
if(~ tree[p].tag)
{
tree[p].tag = __gcd(x, tree[p].ma);
tree[p].ma = tree[p].tag;
return ;
}
}
pushdown(p, nl, nr);
int mid = midf(nl, nr);
if(l <= mid) update_2(nl, mid, DXA(p));
if(mid < r) update_2(mid + 1, nr, DXB(p));
pushup(p, nl, nr);
}
void query_max(int nl, int nr, int p)
{
if(~ tree[p].tag)
{
for(int i = nl; i <= nr; i ++)
printf("%d ", tree[p].tag);
return ;
}
if(nl == nr)
{
printf("%d ", tree[p].ma);
return ;
}
pushdown(p, nl, nr);
int mid = midf(nl, nr);
query_max(nl, mid, DXA(p));
query_max(mid + 1, nr, DXB(p));
}
int main()
{
int T, type;
//scan_d(T);
scanf("%d", &T);
while(T --)
{
//scan_d(n); scan_d(m);
scanf("%d", &n);
pre(1, n, 1);
scanf("%d", &m);
while(m --)
{
//scan_d(type);
scanf("%d %d %d %d", &type, &l, &r, &x);
switch(type)
{
case 1:
update_1(1, n, 1);
break;
case 2:
update_2(1, n, 1);
break;
default:
assert(type == 1 || type == 2);
}
}
query_max(1, n, 1);
puts("");
}
return 0;
}

未解决的问题

文章目录
  1. 1. HDU 4902 Nice boat
    1. 1.1. 题目翻译
    2. 1.2. 题目分析
      1. 1.2.1. 题解
    3. 1.3. AC Code
  • 未解决的问题
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