You are in a reality show, and the show is way too real that they threw into an island. Only two kinds of animals are in the island, the tigers and the deer. Though unfortunate but the truth is that, each day exactly two animals meet each other. So, the outcomes are one of the following
a) If you and a tiger meet, the tiger will surely kill you.
b) If a tiger and a deer meet, the tiger will eat the deer.
c) If two deer meet, nothing happens.
d) If you meet a deer, you may or may not kill the deer (depends on you).
e) If two tigers meet, they will fight each other till death. So, both will be killed.
If in some day you are sure that you will not be killed, you leave the island immediately and thus win the reality show. And you can assume that two animals in each day are chosen uniformly at random from the set of living creatures in the island (including you).
Now you want to find the expected probability of you winning the game. Since in outcome (d), you can make your own decision, you want to maximize the probability.
Input
Input starts with an integer T (≤ 200), denoting the number of test cases.
Each case starts with a line containing two integers t (0 ≤ t ≤ 1000) and d (0 ≤ d ≤ 1000) where tdenotes the number of tigers and d denotes the number of deer.
Output
For each case, print the case number and the expected probability. Errors less than 10-6 will be ignored.
Sample Input
4
0 0
1 7
2 0
0 10
Sample Output
Case 1: 1
Case 2: 0
Case 3: 0.3333333333
Case 4: 1
思路：其实和鹿的数量是没有关系的，因为老虎吃了鹿，它还是活着的，所以人的危险还是存在的，所以只考虑老虎的数量+人的数量就行，如果老虎为奇数个，那么总是会剩下一头的，所以人必死，接着，用乘法原理累乘下两只虎相遇的概率就行了。
代码：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<vector>
#include<algorithm>
#include<set>
#include<sstream>
#include<cstdio>
#include<unordered_map>
#include<unordered_set>
#include<cmath>
#include<climits>
using namespace std;
const int maxn=2e6+9;
int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int T;
cin>>T;
int Case=0;
while(T--)
{
int t,d;
double ans;
scanf("%d%d",&t,&d);
printf("Case %d: ",++Case);
if(t&1)
{
printf("0\n");
}
else
{
ans=1.0;
while(t)
{
ans*=1.0*(t-1)/(t+1);
t-=2;
}
printf("%.7lf\n",ans);
}
}
return 0;
}