CodeForces 1042D

该抖的时候就得抖

CodeForces 1042D Petya and Array

题目描述

给你一个数组,询问区间连续和小于 $x$ 有多少对。

题解

Splay

对于这次的前缀和 $s[i]$ ,在平衡树上求出第 $k$ 大(对 $s[k]$ 从小到大的定义下 )满足 $s[i] + s[k] \geq t$ 。那么第 $k$ 大以前就满足题意了,然后将 $s[i]$ 丢进平衡树里面。

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 <climits>
#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 = 300009;
const int maxm = 300009;
const double pi = acos(-1.0);
const double eps = 1e-6;
const ll inf = 1e16;
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');
}
#define ls(p) t[p].ch[0]
#define rs(p) t[p].ch[1]
struct node
{
int size;
ll v;
int f, ch[2];
node() {}
node(ll val, int fa) {ch[0] = ch[1] = 0; f = fa; size = 1; v = val;}
}t[maxn];
int st[maxn], sz, top, root;
int newnode() {return top ? st[top--] : ++sz;}
int wh(int p) {return t[t[p].f].ch[1] == p;}
void pushup(int p)
{
t[p].size = t[ls(p)].size + t[rs(p)].size + 1;
}
void rotate(int p)
{
int f = t[p].f, g = t[f].f, c = wh(p);
if(g) t[g].ch[wh(f)] = p; t[p].f = g;
t[f].ch[c] = t[p].ch[c ^ 1]; if(t[f].ch[c]) t[t[f].ch[c]].f = f;
t[p].ch[c ^ 1] = f; t[f].f = p;
pushup(f); pushup(p);
}
void Splay(int p, int cur)
{
for(;t[p].f != cur; rotate(p))
if(t[t[p].f].f != cur) rotate(wh(p) == wh(t[p].f) ? t[p].f : p);
if(cur == 0) root = p;
}
void init()
{
sz = 0;
top = 0;
t[0].v = -inf;
root = newnode();
t[root] = node(-inf, 0);
rs(root) = newnode();
t[rs(root)] = node(inf, root);
pushup(root);
}
void insert(int &p, ll val, int fa)
{
if(p == 0) {p = newnode(); t[p] = node(val, fa); return ;}
if(t[p].v > val) insert(ls(p), val, p);
else insert(rs(p), val, p);
pushup(p);
}
int kth_place;
ll qd;
void next_val(int p, ll val)
{
if(! p) return;
if(t[p].v <= val) next_val(rs(p), val);
else
{
if(t[p].v - val <= qd - val) kth_place = p, qd = t[p].v;
next_val(ls(p), val);
}
Splay(p, 0);
}
#undef ls
#undef rs
ll sum[maxn], A[maxn], v;
int main()
{
int n;
scanf("%d %I64d", &n, &v);
sum[0] = 0;
for(int i = 1; i <= n; ++ i) scanf("%I64d", A + i), sum[i] = sum[i - 1] + A[i];
init();
ll ans = 0;
insert(root, 0, 0);
for(int i = 1; i <= n; ++ i)
{
Splay(i, 0);
kth_place = -1;
qd = inf;
next_val(root, sum[i] - v);
//printf("%d %I64d\n", kth_place, qd);
if(!~kth_place) continue;
Splay(kth_place, 0);
ans += t[t[kth_place].ch[1]].size;
insert(root, sum[i], 0);
Splay(sz, 0);
}
printf("%I64d\n", ans);
return 0;
}

树状数组

前缀和的处理丢进树状数组就行。

树状数组代码

文章目录
  1. 1. CodeForces 1042D Petya and Array
    1. 1.1. 题目描述
    2. 1.2. 题解
      1. 1.2.1. Splay
        1. 1.2.1.1. AC Code
      2. 1.2.2. 树状数组
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