Codeforces Round #673 (Div. 2) Problem A

今天的题。
本来打算把比赛坚持打完的，但是因为生病了，还是早点睡吧，把第一题摸了。
题面如下：
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains ai

candies. BThero can cast a copy-paste spell as follows:

He chooses two piles (i,j)

such that 1≤i,j≤n and i≠j
.
All candies from pile i
are copied into pile j. Formally, the operation aj:=aj+ai

is performed.

BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k

candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input

The first line contains one integer T
(1≤T≤500

) — the number of test cases.

Each test case consists of two lines:

the first line contains two integers n

and k (2≤n≤1000, 2≤k≤104
);
the second line contains n
integers a1, a2, …, an (1≤ai≤k

).

It is guaranteed that the sum of n
over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 104

.
Output

For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input

3
2 2
1 1
3 5
1 2 3
3 7
3 2 2

Output

1
5
4

解题的思路还是很简单的，最优策略一定是用最小的值把每个其他数加到刚好小于k，这样的策略是最佳的。
证明从略，以后可能会补充。
代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1005
using namespace std;
int T;
int n,k;
int a[maxn];
int minn,ind;
int ans;
int main(){
  scanf("%d",&T);
  while(T--){
    scanf("%d%d",&n,&k);
    minn=0x3f3f3f3f;
    for(int i=0;i<n;i++){
      scanf("%d",&a[i]);
      if(minn>a[i]){
        ind=i;
        minn=a[i];
      }
    }
    ans=0;
    for(int i=0;i<n;i++){
      if(i!=ind){
        ans+=(k-a[i])/minn;
      }
    }
    printf("%d\n",ans);

  }
  return 0;
}