HDU 5365
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Run
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 957 Accepted Submission(s): 413
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.
Please tell her how many ways can her find.
Two ways are same if the set of chair that they contains are same.
Please tell her how many ways can her find.
Two ways are same if the set of chair that they contains are same.
Input
There are multiply case.
In each case,there is a integer n(1 < = n < = 20)in a line.
In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.
In each case,there is a integer n(1 < = n < = 20)in a line.
In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.
Output
Output the number of ways.
Sample Input
40 00 11 01 1
Sample Output
1
Source
BestCoder Round #50 (div.2)
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/* * In the name of god (^_^) */#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cmath>#include<stdlib.h>#include<map>#include<set>#include<time.h>#include<vector>#include<queue>#include<string>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define eps 1e-8#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LL long long#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))typedef pair<int , int> pii;#define maxn 22int n;struct point{ int x, y;} P[maxn];int dist(point a, point b){ return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);}int D[6];LL ans ;int main(){ while(~scanf("%d", &n)) { ans = 0; for(int i = 1; i <= n; i++) scanf("%d%d", &P[i].x, &P[i].y); for(int i = 1; i + 3 <= n; i++) for(int j = i + 1; j + 2 <= n; j++) for(int t = j + 1; t + 1 <= n; t++) for(int k = t + 1; k <= n; k++) { D[0] = dist(P[i], P[j]); D[1] = dist(P[i], P[t]); D[2] = dist(P[i], P[k]); D[3] = dist(P[j], P[t]); D[4] = dist(P[j], P[k]); D[5] = dist(P[t], P[k]); sort(D, D + 6); int num; for(num = 0; num < 3; num++) { if(D[num] != D[num + 1]) break; } if(num == 3 && D[num + 1] == 2 * D[num] && D[num]) { ans++; } } printf("%I64d\n", ans); } return 0;}/*40 00 00 00 050 11 01 22 12 211 170 00 11 01 10 22 02 210 0*/
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