| 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 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
Source
求满足T(C-T*N) 值最大时的T值, x=对称轴时y最大 ; /*利润最大时取整 ;*/ 关键是读题;
#include <cstdio> typedef long long LL; int main() { int t, Q=1; scanf("%d", &t); while(t--) { LL N, C; scanf("%lld%lld", &N, &C); if(N==0||C==0) // 0 和 开口向下的一元二次(顶点在原点)方程 ; { printf("Case %d: %lld ", Q++, 0); continue; } LL T=C/2/N; LL rec1=T*(C-T*N); LL rec2=(T+1)*(C-(T+1)*N); LL rec=T; if(rec1<rec2) rec=T+1; printf("Case %d: %lld ", Q++, rec); } return 0; }