第六届福建省大学生程序设计竞赛 Problem A Super Mobile Charger【模拟】

素颜马尾好姑娘i 2022-07-31 09:56 140阅读 0赞

**Problem A Super Mobile Charger**

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Accept: 209 Submit: 391

Time Limit: 1000 mSec Memory Limit : 32768 KB

## ![prodesc.gif][] Problem Description ##

While HIT ACM Group finished their contest in Shanghai and is heading back Harbin, their train was delayed due to the heavy snow. Their mobile phones are all running out of battery very quick. Luckily, zb has a super mobile charger that can charge all phones.

There are N people on the train, and the i-th phone has p\[i\] percentage of power, the super mobile charger can charge at most M percentage of power.

Now zb wants to know, at most how many phones can be charged to full power. (100 percent means full power.)

## ![prodesc.gif][] Input ##

The first line contains an integer T, meaning the number of the cases (1 <= T <= 50.).

For each test case, the first line contains two integers N, M(1 <= N <= 100,0 <= M <= 10000) , the second line contains N integers p\[i\](0 <= p\[i\] <= 100) meaning the percentage of power of the i-th phone.

## ![prodesc.gif][] Output ##

For each test case, output the answer of the question.

## ![prodesc.gif][] Sample Input ##

3 10

100 99 90

3 1000

0 0 0

## ![prodesc.gif][] Sample Output ##

题目大意：

有电能为M的充电宝，和N个剩余电量的手机，问充电宝最多能把几个手机充到满电。

解题思路：

排序从用数组存一下和100的差值，从小到大排序。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std; 
    int map[10000];
    bool cmp(int n,int y)
    {
    	return n<y;
    }
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		int n;
    		int m;
    		scanf("%d%d",&n,&m);
    		int i,j;
    		for(i=0;i<n;i++)
    		{
    			int yy;
    			scanf("%d",&yy);
    			map[i]=100-yy;
    		}
    		int ans=0;
    		sort(map,map+n,cmp);
    		for(i=0;i<n;i++)
    		{
    			if(m-map[i]>=0)
    			{
    				ans++;
    				m-=map[i];
    			}
    		}
    		printf("%d\n",ans);
    	}
    	return 0; 
    }

[prodesc.gif]: http://acm.fzu.edu.cn/image/prodesc.gif