poj 2291水题

http://poj.org/problem?id=2291

线段树做累了来题水题

Rotten Ropes

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4899 Accepted: 3184

Description

Suppose we have n ropes of equal length and we want to use them to lift some heavy object. A tear-off weight t is associated to each rope, that is, if we try to lift an object, heavier than t with that rope, it will tear off. But we can fasten a number of ropes to the heavy object (in parallel), and lift it with all the fastened ropes. When using k ropes to lift a heavy object with weight w, we assume that each of the k ropes, regardless of its tear-off weight, is responsible for lifting a weight of w/k. However, if w/k > t for some rope with tear-off weight of t, that rope will tear off. For example, three ropes with tear-off weights of 1, 10, and 15, when all three are fastened to an object, can not lift an object with weight more than 3, unless the weaker one tears-off. But the second rope, may lift by itself, an object with weight at most 10. Given the tear-off weights of n ropes, your task is to find the weight of the heaviest object that can be lifted by fastening a subset of the given ropes without any of them tearing off.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 1000) which is the number of ropes. Following the first line, there is a single line containing n integers between 1 and 10000 which are the tear-off weights of the ropes, separated by blank characters.

Output

Each line of the output should contain a single number, which is the largest weight that can be lifted in the corresponding test case without tearing off any rope chosen.

Sample Input

2310 1 15210 15

Sample Output

2020

Source

Tehran 2003

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn =1010;
int a[maxn];
const int INF = 0x3f3f3f3f;


int cmp(int a,int b){
    return a>b;
}


int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        sort(a,a+n,cmp);
//        for(int i=0;i<n;i++){
//            printf("a[i]=%d\n",a[i]);
//        }
        int maxx=-INF;
        for(int i=0;i<n;i++){
            if(a[i]*(i+1)>maxx){
                maxx=a[i]*(i+1);
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}