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
#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; }