| Time Limit: 3 second(s) | Memory Limit: 32 MB |
You are given a rectangular grid of height H and width W. A problem setter wants to draw a pair of circles inside the rectangle so that they touch each other but do not share common area and both the circles are completely inside the rectangle. As the problem setter does not like precision problems, he also wants their centers to be on integer coordinates and their radii should be positive integers as well. How many different ways can he draw such pair of circles? Two drawings are different from each other if any of the circles has different center location or radius.
Input
Input starts with an integer T (≤ 500), denoting the number of test cases.
Each case starts with a line containing two integers H and W (0 < H, W ≤ 1000).
Output
For each case, print the case number and the number of ways of drawing such pairs of circles maintaining the mentioned constraints. Each output will fit into a 64-bit signed integer.
Sample Input |
Output for Sample Input |
|
5 |
Case 1: 1 Case 2: 2 Case 3: 6 Case 4: 16 Case 5: 496 |
Note
For case 3, the possible results are:
链接：http://lightoj.com/volume_showproblem.php?problem=1366
枚举两个圆心的相对坐标，然后枚举其中的一个半径。
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013-10-18 17:27:32 4 File Name :E:2013ACM专题强化训练计算几何LightOJ1366.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 21 int main() 22 { 23 //freopen("in.txt","r",stdin); 24 //freopen("out.txt","w",stdout); 25 int T; 26 int w,h; 27 int iCase = 0; 28 scanf("%d",&T); 29 while(T--) 30 { 31 iCase++; 32 scanf("%d%d",&w,&h); 33 long long ans = 0; 34 //枚举两个圆心的相对坐标 35 for(int i = 0;i <= w/2;i++) 36 for(int j = 0;j <= h/2;j++) 37 { 38 if(i == 0 && j == 0)continue; 39 int tmp = sqrt(i*i + j*j); 40 if(tmp*tmp != i*i + j*j)continue; 41 //枚举其中一个圆的半径 42 for(int k = 1;k < tmp;k++) 43 { 44 int y1 = min(-k,j-(tmp-k)), y2 = max(k,j+(tmp-k)); 45 int x1 = min(-k,i-(tmp-k)), x2 = max(k,i+(tmp-k)); 46 int x = x2-x1; 47 int y = y2-y1; 48 if(x > w || y > h)continue; 49 long long tt = (w - x + 1)*(h - y + 1); 50 if(i*j)tt*= 2; 51 ans += tt; 52 } 53 } 54 printf("Case %d: ",iCase); 55 cout<<ans<<endl; 56 } 57 return 0; 58 }