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;
}