POJ 3911：Internet Service Providers

Internet Service Providers

Time Limit: 2MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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 - TN). 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.

Input

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.

Output

For each data set the program computes the value of T_optim according to the problem instance that corresponds to the data set. The result is printed on the standard output from the beginning of a line. There must be no empty lines on the output.

Sample Input

1 0
0 1
4 3
2 8
3 27
25 1000000000

Sample Output

0
0
0
2
4
20000000

求一个一元二次方程取极值时x的值。-b/(2*a)。。。

因为程序里面除法可能消掉了一部分小数，所以还要判断x+1时的情况，比较一下再输出。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	long long n,c;
	while(scanf("%lld%lld",&n,&c)!=EOF)
	{
		if(n==0)
		{
			cout<<0<<endl;
			continue;
		}
		else
		{
			long long x=c/(2*n);
			long long y=x+1;//一开始竟写成了减1。。。
			cout<<(x*(c-n*x)<y*(c-n*y)?y:x)<<endl;
		}
	}
	return 0;
}

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