poj2291

Rotten Ropes Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4150 Accepted: 2697 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

 

水过

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int cas;
    int n;
    int tear_weight[1000];
    int max_weight;

    cin>>cas;
    while(cas--)
    {
        cin>>n;
        for(int i=0; i<n; i++)
            cin>>tear_weight[i];
        sort(tear_weight,tear_weight+n);
        max_weight=0;
        for(int i=0; i<n; i++)
        {
           if((n-i)*tear_weight[i]>max_weight)
                max_weight=(n-i)*tear_weight[i];
        }
        cout<<max_weight<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/eric-blog/archive/2011/06/02/2068436.html