【HDU 5194】DZY Loves Balls

相关链接

题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=5194
中文题面:http://blog.csdn.net/u014086857/article/details/44724335

解题报告

这题主要就是用到一个“期望的可加性”吧!
什么是可加性呢?

期望的和,等于和的期望

这里不需要期望之间相互独立
证明的话,你可以参见这里:http://blog.csdn.net/grooowing/article/details/45000205

于是这题我们分开考虑每一个位置出现01的概率就好了
于是推一推式子发现答案至于n,m相关,于是写一个gcd就好了

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;

inline int read() {
	char c=getchar(); int f=1,ret=0;
	while (c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
	while (c<='9'&&c>='0') {ret=ret*10+c-'0';c=getchar();}
	return ret * f;
}

int GCD(int x, int y) {
	return y? GCD(y, x % y): x;
} 

int main() {
	for (int n,m,gcd;~scanf("%d%d",&n,&m);) {
		gcd = GCD(n*m, n+m);
		printf("%d/%d\n", n*m/gcd, (n+m)/gcd);
	}
	return 0;
}

2 thoughts to “【HDU 5194】DZY Loves Balls”

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