数学规律，一元二次方程求最大值。

E - E

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit  Status

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 optimalT 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：求一元二次方程T*(C-N*T)取得最大值时，T的值是多少。

ac代码：

//坑死，关键还是对题目的理解。 
#include
#include
#include
int main(){
	int T;
	scanf("%d",&T);
	for(int i=1;i<=T;i++){
		long long N,C;
		scanf("%lld%lld",&N,&C);
		if(N==0){
			printf("Case %d: %d\n",i,0);
			continue ;
		}
		long long n,m;
		float mid;
		mid=(float)C/2/N;
		n=floor(mid);//向下取整 
		m=ceil(mid);//向上取整 
		long long ans;
		long long temp1=n*(C-N*n),temp2=m*(C-N*m);
		if(temp1>=temp2){//尽量减少使用实数进行大小的判断。 
			ans=n;
		}
		else
			ans=m;
		printf("Case %d: %lld\n",i,ans);
	}
	return 0;
}