LightOJ Problem 1275 【解方程水题】

Internet Service Providers

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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 NISPs 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

求一元二次函数的最大值点，只不过因为解是整数，四舍五入会有偏差，要多多判断解的前一个值。

#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    double n, c;
    scanf("%d", &t);
    int cnt = 0;
    while (t--) {
        scanf("%lf%lf", &n, &c);
        if(n == 0){
            printf("Case %d: 0
", ++cnt); continue;
        }
        long long ans = c/(2*n);

        if (ans*(c-ans*n) < (ans+1)*(c-(ans+1)*n))
            ans++;
        printf("Case %d: %lld
", ++cnt, ans);
    }
    return 0;
}