POJ2291——Rotten Ropes

时间：2015-06-08 21:06:32 阅读：169 评论：0 收藏：0 [点我收藏+]

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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

Sample Output

20
20

Source

Tehran 2003

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
bool cmp(int a,int b)
{
    return a>b;
}
int a[1010];
int main()
{
    int T;
    int n;
    scanf("%d",&T);
    while(T--){
            scanf("%d",&n);
    for(int i = 1; i <= n ; i++){
            scanf("%d",&a[i]);
    }
    sort(a+1,a+n+1,cmp);
    int max1 = 0;
    for(int i = 1 ; i <= n ; i++){
            max1 = max(max1,a[i]*i);
    }
    printf("%d\n",max1);
    }
    return 0;
}

POJ2291——Rotten Ropes

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原文地址：http://www.cnblogs.com/zero-begin/p/4561703.html

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