hdu 1017 A Mathematical Curiosity

# 移动开发 2024-05-05 15:32 0 28 来源：云博客

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43299 Accepted Submission(s): 13943

Problem Description

Given two integers n and m, count the number of pairs of integers (a,b) such that 0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

You will be given a number of cases in the input. Each case is specified by a line containing the integers n and m. The end of input is indicated by a case in which n = m = 0. You may assume that 0 < n <= 100.

Output

For each case, print the case number as well as the number of pairs (a,b) satisfying the given property. Print the output for each case on one line in the format as shown below.

Sample Input

1 10 1 20 3 30 4 0 0

Sample Output

Case 1: 2 Case 2: 4 Case 3: 5

Source

East Central North America 1999, Practice



//几何的简单模板题
 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 #define INF 0x7f7f7f 8 int n; 9 struct Point10 {11 double x,y;12 }p[100001];13 bool cmp(Point a,Point b)14 {15 if(a.x==b.x)16  {17 return a.y<b.y;18  }19 return a.x<b.x;20 }21 double dis(Point a,Point b)22 {23 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));24 }25 double getMin(int i)26 {27 double d=999999,di=0;28 for(int j=i+1;j<n;j++)29  {30 di=dis(p[i],p[j]);31 if(d>di)d=di;32 else break;33  }34 return d;35 }36 int main()37 {38 int i,j,k;39 double ans,f;40 while(scanf("%d",&n)!=EOF)41  {42 if(!n)break;43 for(i=0;i<n;i++)44  {45 scanf("%lf%lf",&p[i].x,&p[i].y);46  }47 sort(p,p+n,cmp);48 ans=INF;49 for(i=0;i<n-1;i++)50  {51 f=getMin(i);52 if(ans>f)ans=f;53  }54 printf("%.2lf\n",ans/2.0);55  }56 return 0;57 }