LIGHTOJ-1275 - Internet Service Providers

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1275 - Internet Service Providers
Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for Sample Input

6

1 0

0 1

4 3

2 8

3 27

25 1000000000

Case 1: 0

Case 2: 0

Case 3: 0

Case 4: 2

Case 5: 4

Case 6: 20000000

 



解一个一元二次方程,求出什么时候他的值最大,就是求对称轴的,但是注意x只能为整数且非负!



#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	long long a,b,m,n,s,cot=1;
	scanf("%lld",&n);
	while(n--)
	{
		scanf("%lld%lld",&a,&b);
		if(!a||!b)
		{
			printf("Case %lld: 0\n",cot++);
			continue;
		}
			double t=b*1.0/(2*a);
			if(t<=0)
			printf("0\n");
			else
			{
				if(t<=(int)t+0.5)
				s=(int)t;
				else
				s=(int)t+1;
			}
		printf("Case %lld: %lld\n",cot++,s);
	}
	return 0;
}


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