Super Mobile Charger

 Super Mobile Charger

Accept: 8 Submit: 24

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

23

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）   http://acm.fzu.edu.cn/problem.php?pid=2212
解题思路：比赛的大水题，由于自己第一次看错题目，第二次写排序时出问题导致wa了。大概思路就是先把n个数进行排序由大到小。然后在开始100-a[i]的差和m比较,如果m大于0的话就一直重复操作

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<iostream>
#include<algorithm>
using namespace std;

int  cmp(int a,int b)
{
return a>b;
}
int main()
{
int t;
cin>>t;
while(t--)
{
int n,m;
cin>>n>>m;

int a[101]={0};
for(int i=0;i<n;i++)
{

cin>>a[i];

}
sort(a,a+n,cmp);
int tp=0;
for(int i=0;i<n;i++)
{
m=m-(100-a[i]);
if(m>=0)
{
tp++;

}
}

cout<<tp<<endl;

}
return 0;
}