FZOJ--2212--Super Mobile Charger（水题）

最新推荐文章于 2018-09-09 15:47:20 发布

转载最新推荐文章于 2018-09-09 15:47:20 发布 · 69 阅读

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原文链接：http://www.cnblogs.com/playboy307/p/5273592.html

本文介绍了一种名为超级移动充电器的问题解决方案。该方案通过算法确定在给定的充电量下，最多能将多少部手机充满电。文章提供了一个具体的编程实现案例，包括输入输出示例及完整的C++代码。

Problem 2212 Super Mobile Charger

Accept: 3 Submit: 11

Time Limit: 1000 mSec Memory Limit : 32768 KB

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.)

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.

Output

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

Sample Input

23 10100 99 903 10000 0 0

Sample Output

Source

第六届福建省大学生程序设计竞赛-重现赛（感谢承办方华侨大学）

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int num[100100];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m;
		memset(num,0,sizeof(num));
		scanf("%d%d",&n,&m);
		int x;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			num[i]=100-x;
		}
		sort(num,num+n);
		int ans=0;
		for(int i=0;i<n;i++)
		{
			if(m>=num[i])
			{
				m-=num[i];
				ans++;
			}			
		}
		printf("%d\n",ans);
	}
	return 0;
}

转载于:https://www.cnblogs.com/playboy307/p/5273592.html