LightOJ 1275：Internet Service Providers
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
求解一元二次方程 我们知道 -ax^2+bx+c 最大值是在x= -b/(2a)处取得
<pre name="code" class="cpp">#include<cstdio>
int main()
{
long long t,cut=0;
scanf("%lld",&t);
while(t--)
{
cut++;
long long n,c;
scanf("%lld%lld",&n,&c);
printf("Case %lld: ",cut);
if(n==0)
{
printf("0\n");
continue;
}
long long a=(c/n)/2;//这里结果可能为小数 因为定义为int型结果就只取整数部分 那么结果就不一定是最大了 需要将x向左或右移一位
long long b=a+1;
printf("%lld\n",a*(c-a*n)>=b*(c-b*n)?a:b);
}
return 0;
} 