LightOJ 1275：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

求解一元二次方程    我们知道 -ax^2+bx+c 最大值是在x=  -b/(2a)处取得

<pre name="code" class="cpp">#include<cstdio>int main(){long long t,cut=0;scanf("%lld",&t);while(t--){cut++;long long n,c;scanf("%lld%lld",&n,&c);printf("Case %lld: ",cut);if(n==0){printf("0\n");continue;}      long long a=(c/n)/2;//这里结果可能为小数 因为定义为int型结果就只取整数部分  那么结果就不一定是最大了 需要将x向左或右移一位     long long b=a+1;  printf("%lld\n",a*(c-a*n)>=b*(c-b*n)?a:b);}return 0; }