When I was young, my father is a senior gaming enthusiast. One day, we saw a old man in the street . He had a dice and played with other people . Every turn the gambler gives k RMB to the old man and throw the dice . If the point is 1,2 or 3fi he will win 2k RMB back fiotherwise he will get nothing . My father told me, “I can win all his money by the following strategy”. “Each turn I bet on 1 RMB first If I lose I will bet on 2 RMB. If I still lose, I will bet on 4,8,16,… and so on, until I win. And start to bet on 1 RMB, do the same thing again .” “If I don’t have enough money to bet, I will bet on all my money.” Now the question is, if the dice is even, my father has n RMB, the old man has m RMB, they stop until one of them lose all his money, what’s the probability of my father’s victory . Input The input contains multiple test cases.(No more than 20) In each test case: The only line contains two numbers n, m. (0 ≤ n, m ≤ 2000000), indicate my father’s money and the old man’s . We guarantee max(n, m) ≥ 1. Output For each test case, print the answer in five decimal . Sample Input 1 0 3 3 Sample Output 1.00000 0.50000

• 解題思路：直接輸出n/(m+n)即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool SUBMIT = 1;
int main()
{
	double n,m;
	while(~scanf("%lf %lf",&n,&m)){
	//cout<<n<<m<<endl;
	printf("%.5lf\n",n/(n+m));
	}
	return 0;
}