Description

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 t denotes 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
题目很好理解，我这英语渣都看得懂。。。。
主要是要忽略掉deer的影响，我之前没想到。。。
然后wa了2发。。。
看了看别人的都写的循环，其实没必要。。。
如下
t个老虎，如果是奇数，那你必死；
偶数的话活下来的情况就是老虎都自相残杀啦；
t/(t+1)*（t-1）/t=（t-1)/(t+1)
然后继续连乘（t-1)/(t+1)*(t-3)/(t-1)~~~1/3;
约分后 ans=1/（t+1)
这样快多啦。。。
ac代码`

include

include

include

include

include

using namespace std;

define debug(x) cout<<#x<<’=’<

define ya(x) scanf(“%d”,&x)

define Cas(x) printf(“Case %d: “,x)

define wen(x) printf(“%d\n”,x)

define FOR(T) for(int cas=1;cas<=T;cas++)

int main()
{
int T;
//freopen(“input.txt”,”r”,stdin);
ya(T);
FOR(T)
{
int t,d;
cin>>t>>d;
Cas(cas);
if(t%2==0)
printf(“%lf\n”,1.0/(t+1));
else printf(“0\n”);
}
return 0;
}