HihoCoder 1269 优化延迟（二分、优先队列）

题意:

不是很好解释，反正中文题目就不写了

分析:

可以发现缓存越大，我们就越能拿出最大的那个数，然后让它的贡献变小
所以这里有单调递减性，随着缓存变大
然后直接二分缓存大小，优先队列模拟即可

代码:

////  Created by TaoSama on 2016-03-06//  Copyright (c) 2016 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n, a[N];LL Q;bool check(int x) {    priority_queue<int> q;    LL ret = 0;    int j = 1;    for(int i = 1; i <= n; ++i) {        if(q.size() == x) {            int cur = q.top(); q.pop();            ret += cur * j++;        }        q.push(a[i]);    }    while(q.size()) {        int cur = q.top(); q.pop();        ret += cur * j++;    }    return ret <= Q;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    scanf("%d%lld%", &n, &Q);    for(int i = 1; i <= n; ++i) scanf("%d", a + i);    int l = 1, r = 100000;    while(l <= r) {        int m = l + r >> 1;        if(check(m)) r = m - 1;        else l = m + 1;    }    if(l > 100000) l = -1;    printf("%d\n", l);    return 0;}