poj 3911（Internet Service Providers）

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 T_optim, 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 T_optim are integer numbers.

输入格式
Each data set corresponds to an instance of the problem above and contains two integral numbers - N and C - with values in the range from 0 to 10^9. The input data are separated by white spaces, are correct, and terminate with an end of file. For each data set the program computes the value of T_optim according to the problem instance that corresponds to the data set.

输出格式
The result is printed on the standard output from the beginning of a line. There must be no empty lines on the output. An example of input/output is shown below.

样例输入
1 0
0 1
4 3
2 8
3 27
25 1000000000

样例输出
0
0
0
2
4
20000000

水题，解出方程，不过要使T尽量小

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e6+5;
int main() {
    LL n,c;
    int T;
     scanf("%d",&T);
    int cas=0;
    while(T--)
    {
    scanf("%lld %lld",&n,&c);
    
        if(n==0||c==0) printf("Case %d: 0\n",++cas);
        else 
        {
            LL t=c/(2*n);
            //cout<<t<<endl;
            //cout<<t*(c-t*n)<<" "<<(t+1)*(c-(t+1)*n)<<endl;
            if(t*(c-t*n)>=(t+1)*(c-(t+1)*n)) 
                printf("Case %d: %lld\n",++cas,t);
            else 
               printf("Case %d: %lld\n",++cas,t+1);;
        }
    }
    return 0;
}