[codeforces 1304B] Cow and Friend 除+模

Codeforces Round #621 (Div. 1 + Div. 2)

[codeforces 1304B] Cow and Friend   除+模

总目录详见https://blog.csdn.net/mrcrack/article/details/103564004

在线测评地址https://codeforces.ml/contest/1307/problem/B

Problem	Lang	Verdict	Time	Memory
B - Cow and Friend	GNU C++11	Accepted	46 ms	200 KB

举个例子

Input3:
1
1 10
3
Output3:
4

构图如下

再多举几个例子，弄明白了，算法也就基本明白了。

提一句，题中的思路是给手工算算的，要给计算机 算，还得读者自己再归纳。

Input1:
1
1 10
1
Output1:
10

Input2:
1
1 10
2
Output2:
5

Input3:
1
1 10
3
Output3:
4

Input4:
1
1 10
4
Output4:
3

Input5:
1
1 10
5
Output5:
2

Input6:
1
1 10
6
Output6:
2

Input7:
1
1 10
7
Output7:
2

Input8:
1
1 10
8
Output8:
2

Input9:
1
1 10
9
Output9:
2

Input10:
1
1 10
10
Output10:
1

Input11:
1
1 10
11
Output11:
2

Input12:
1
1 10
12
Output12:
2

Input13:
1
1 10
100
Output13:
2

AC代码如下

#include <stdio.h>
int a[100010];
int min(int a,int b){
	return a<b?a:b;
}
int main(){
	int t,n,x,i,mn;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&x),mn=1000000010;
		for(i=1;i<=n;i++)scanf("%d",&a[i]);
		for(i=1;i<=n;i++){
			if(a[i]<=x)a[i]=x/a[i]+(x%a[i]>0);
			else a[i]=2;//a[i]>x
			mn=min(mn,a[i]);
		}
		printf("%d\n",mn);
	}
	return 0;
}