标签：acm算法
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Run Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description AFA is a girl who like runing.Today,he download an app about runing .The app can record the trace of her runing.AFA will start runing in the park.There are many chairs in the park,and AFA will start his runing in a chair and end in
this chair.Between two chairs,she running in a line.she want the the trace can be a regular triangle or a square or a regular pentagon or a regular hexagon.
Input There are multiply case.
Output Output the number of ways.
Sample Input 4 0 0 0 1 1 0 1 1
Sample Output 1
Source
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#include <cstdio>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
int a[10];
int distance1(int x,int y)
{
return x*x+y*y;
}
int main()
{
int n;
int x,y;
while(cin>>n)
{
int i,j,k,l;
int x[21];
int y[21];
int res=0;
memset(a,0,sizeof(a));
for(i=0; i<n; i++)
cin>>x[i]>>y[i];
for(i=0; i<n; i++)
for(j=i+1; j<n; j++)
for(k=j+1; k<n; k++)
for(l=k+1; l<n; l++)
{
a[0]=distance1(x[i]-x[j],y[i]-y[j]);
a[1]=distance1(x[i]-x[k],y[i]-y[k]);
a[2]=distance1(x[i]-x[l],y[i]-y[l]);
a[3]=distance1(x[j]-x[k],y[j]-y[k]);
a[4]=distance1(x[j]-x[l],y[j]-y[l]);
a[5]=distance1(x[k]-x[l],y[k]-y[l]);
sort(a,a+6);
if(a[0]==a[1]&&a[1]==a[2]&&a[2]==a[3]&&a[4]==a[5])
res++;
}
cout<<res<<endl;
}
}
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标签：acm算法
原文地址：http://blog.csdn.net/sky_miange/article/details/47684255
