POJ2291——Rotten Ropes

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

2
3
10 1 15
2
10 15

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;
}