Light OJ 1275 Internet Service Providers 【二次函数 水】

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

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

 

恩，题目大意就是说，求使得 t * ( c - t * n ) 最大的 t 的最小整数值，当 n 不为0 时二次函数求对称轴，为 0 时 无最大值为0 

<span style="font-family:Comic Sans MS;">#include <iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;LL n,c;LL f(LL x){    return x*(c-x*n);}int main(){    int t,cnt=0;    LL ans;    scanf("%d",&t);    while(t--)    {        scanf("%lld%lld",&n,&c);        printf("Case %d: ",++cnt);        if(n==0)        {            ans=0;            printf("%lld\n",ans);            continue;        }        if(c%(2*n)==0)            ans=c/(2*n);        else        {            ans=c/(2*n);            LL tem=ans+1;            if(f(ans)<f(tem))                ans=tem;        }        printf("%lld\n",ans);    }    return 0;}</span>