A

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

题目大意：在火车上需要给手机充电，现在有一个容量为power的充电宝给手机充电问最多能充多少个？

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include<set>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
int a[105];
bool cmp(int x,int y)
{
    return x<y;
}
int main()
{
    int repeat;
    while(scanf("%d",&repeat)!=EOF)
    {
        while(repeat--)
        {
            int num,power;
            int count1=0;
            scanf("%d%d",&num,&power);
            for(int i=0;i<num;i++)
                scanf("%d",&a[i]);
            sort(a,a+num,cmp);//这里就是对那些手机的剩余电进行排序
            for(int i=num-1;i>=0;i--){
                if(a[i]==100)
                    count1++;
                else
                if((power+a[i])>=100){
                    count1++;
                    power=power-(100-a[i]);
                }
            }
            printf("%d\n",count1);
    }
}
    return 0;
}