Codeforces673回合區域二第一題

Copy-paste

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
— Hey folks, how do you like this problem?
— That’ll do it.

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
Note
In the first test case we get either a=[1,2] or a=[2,1] after casting the spell for the first time, and it is impossible to cast it again.
理論上這應該算一個入門級的貪心題目……確實挺簡單的。

#include <iostream>
#include <algorithm>
using namespace std;
int candies[2000];
int main()
{
	int T;
	int i, j;
	cin >> T;
	while (T--)
	{
		int ans = 0;
		int n, k;
		cin >> n >> k;
		for (i = 1; i <= n; i++)
			cin >> candies[i];
		sort(candies + 1, candies + i);
		j = 0;
		for (i = 2; i <= n; i++)
		{
			if (k > candies[i])
				ans += (k - candies[i]) / candies[1];
		}
		cout << ans << endl;
	}
}