Internet Service Providers

Description

A group of N Internet Service Provider companies (ISPs) use a private communication channel that has a maximum capacity of C traffic units per second. Each company transfers T traffic units per second through the channel and gets a profit that is directly proportional to the factor T(C - T*N). The problem is to compute the smallest value of T that maximizes the total profit the N ISPs can get from using the channel. Notice that N, C, T, and the optimal T are integer numbers.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with a line containing two integers N and C (0 ≤ N, C ≤ 10⁹).

Output

For each case, print the case number and the minimum possible value of T that maximizes the total profit. The result should be an integer.

Sample Input

6

1 0

0 1

4 3

2 8

3 27

25 1000000000

Sample Output

Case 1: 0

Case 2: 0

Case 3: 0

Case 4: 2

Case 5: 4

Case 6: 20000000

题意：求获得一元二次方程的最大值的跟，

思路：当x=-b/(2*a)时，值最大，但不要忘记，这个数最大值的解可能为小数，但是用除法求的是整数，答案也是求的整数，所以在求一个x+1的值与其进行比较，取最大；

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define size 1000000using namespace std;int main(){int n,c,l=1,t;scanf("%d",&t);while(t--){    long long ans=0,k,h;    scanf("%d%d",&n,&c);printf("Case %d: ",l++);if(n==0){printf("0\n");continue;}k=c/(2*n);h=k+1;ans=k*(c-k*n)>=h*(c-h*n)?k:h;printf("%lld\n",ans);}return 0;}