2291 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

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/*给出n根承重不同的绳子，选出几根使得承重最大且没有绳子断裂*/#include<iostream>#include<algorithm>using namespace std;bool cmp(int a, int b){    return a > b;}int main(){    int t;    cin >> t;    while (t--)    {        int n;        int w[10000];        int max[1000] = { 0 };        cin >> n;        for (int i = 0; i < n; i++)        {            cin >> w[i];        }        sort(w, w + n, cmp);        for (int i = 0; i < n; i++)        {            max[i] = (i + 1)*w[i];        }        sort(max, max + n, cmp);        cout << max[0] << endl;    }    return 0;}