E - Internet Service Providers_g-internet service providers-优快云博客

本文介绍了一个关于多个互联网服务提供商(ISPs)如何通过调整其在共享通信通道上的流量传输来最大化总利润的问题。该通道具有固定的容量限制，每家公司都会产生与其传输量相关的利润。文章提供了一种算法解决方案，用于计算在满足所有条件的情况下，使得总利润最大化的最小传输单位。

E - Internet Service Providers

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit

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 - 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

题意：求二次方程最大值时x的最小值，x为整数。
思路：图像法：由实数范围内的二次方程判断，由 (0 ≤ N, C ≤ 109)，图像开口向下，取得最大值处为-b/2a,求导也可推出。
失误：写了一遍，测试数据时有错误，经检验没考虑n=0的情况，又看了一边题，有看错了t的范围；最后还应考虑t<=c但-c/2n

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;

int main()
{
    __int64 t,n,c,x,k=0;
    cin>>t;

    while(t--)
    {
        cin>>n>>c;

        cout<<"Case "<<++k<<": ";

        if(n==0)
        {
             cout<<"0"<<endl;//n=0,就是0个公司，怎样都没钱 
        }

        else
        {
             x=c/2/n;//n!=0

             if(-n*x*x+c*x>=-n*(x+1)*(x+1)+c*(x+1))
             cout<<x<<endl;//如果x较大，怕越界，上式应该简化
             else
             cout<<x+1<<endl; 
        }

    }
    return 0;
 }