| 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; }