1275 - Internet Service Providers

1275 - Internet Service Providers

PDF (English)

Statistics

Forum

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

思路：求导加二分

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<stdlib.h>
 5 #include<string.h>
 6 #include<math.h>
 7 #include<queue>
 8 #include<set>
 9 #include<stack>
10 #include<map>
11 #include<set>
12 using namespace std;
13 typedef long long LL;
14 LL ask(LL n,LL m, LL z)
15 {
16     return n*(z-n*m);
17 }
18 int main(void)
19 {
20     int i,j,k;
21     scanf("%d",&k);
22     int s;
23     for(s=1; s<=k; s++)
24     {
25         LL x,y;
26         scanf("%lld %lld",&x,&y);
27         LL l=-y;
28         LL r=y;
29         LL ac=0;
30         while(l<=r)
31         {
32             LL mid=(l+r)/2;
33             if(y-2*mid*x<=0)
34             {
35                 ac=mid;
36                 r=mid-1;
37             }
38             else l=mid+1;
39         }
40         LL ck=ac-1;
41         LL sum1=ask(ac,x,y);
42         LL sum2=ask(ac-1,x,y);
43         if(sum2>=sum1)
44             ac=ck;
45         printf("Case %d:",s);
46         printf(" %lld\n",ac);
47     }
48     return 0;
49 }