LightOJ 1275 Internet Service Providers

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

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;
}