LightOJ 1275 Internet Service Providers

最新推荐文章于 2019-02-13 23:28:31 发布

A_ice_

最新推荐文章于 2019-02-13 23:28:31 发布

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本文链接：https://blog.youkuaiyun.com/A_ice_/article/details/50242537

本文探讨了N个互联网服务提供商如何通过优化每秒传输的流量单位数T，来最大化利用共享通信通道的总利润。通过分析输入参数N、C和T之间的关系，本文提供了解决方案，确保在不超过通道最大容量C的情况下，各公司能够获得最大的总收益。

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1275 - Internet Service Providers

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

1 0

0 1

4 3

2 8

3 27

25 1000000000

Case 1: 0

Case 2: 0

Case 3: 0

Case 4: 2

Case 5: 4

Case 6: 20000000

看得懂题就是水题的题……

然而看题看了很久才懂OTZ

#include<stdio.h>
int main()
{
	int t,i;
	long int n,c,T;
	double a,j;
	while(~scanf("%d",&t))
	{
		for(i=1;i<=t;i++)
		{
			scanf("%ld%ld",&n,&c);
			if(n==0)
			{
				printf("Case %d: 0\n",t);
			}
			else
			{
				a=c/2.0/n;
				j=a-(int)a;
				if(j>=0.5)
					T=(int)a+1;
				else
				
					T=(int)a;
				printf("Case %d: %ld\n",t,T);
			}
		}
	}
	return 0;
}