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LightOJ 1275 - Internet Service Providers

时间:2015-08-16 15:17:26      阅读:126      评论:0      收藏:0      [点我收藏+]

标签:lightoj   c语言   acm   技巧   

      1275 - Internet Service Providers
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 Ctraffic units per second. Each company transfers T traffic units persecond through the channel and gets a profit that is directly proportional tothe factor T(C - T*N). The problem is to compute the smallest value of Tthat maximizes the total profit the N ISPs can get from using thechannel. Notice that N, C, T, and the optimal T are integernumbers.

Input

Input starts with an integer T (≤ 20),denoting the number of test cases.

Each case starts with a line containing two integers Nand C (0 ≤ N, C ≤ 109).

Output

For each case, print the case number and the minimumpossible value of T that maximizes the total profit. The result shouldbe an integer.

Sample Input

Output for Sample Input

6

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



一元二次方程求极值,最值的坐标为(-b/2*a,(4ac-b^2)/4a),此题仅仅求出取最值的横坐标即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;
int main()
{
    LL n,m,c,i,x,y;
    LL cla;
    scanf("%lld",&cla);
    for(int gr=1; gr<=cla; gr++)
    {
        printf("Case %lld: ",gr);
        scanf("%lld %lld",&n,&c);
        if(n==0||c==0)
        {
            printf("0\n");
            continue;
        }
        printf("%lld\n",(LL) (1.0*c/(2*n)+0.499999 ) );//注意精度问题+0.499999
    }
    return 0;
}


版权声明:本文为博主原创文章,未经博主允许不得转载。

LightOJ 1275 - Internet Service Providers

标签:lightoj   c语言   acm   技巧   

原文地址:http://blog.csdn.net/grit_icpc/article/details/47701987

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