codeforces C. Triangle

C. Triangletime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere is a right triangle with legs of length a and b.Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to outputthe appropriate coordinates of vertices.InputThe first line contains two integers a, b (1 ≤ a, b ≤ 1000),separated by a single space.OutputIn the first line print either "YES" or "NO" (without the quotes)depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 intheir absolute value.Sample test(s)input

1 1

output

NO

input

5 5

output

YES2 15 5-2 4

input

5 10

output

YES
-10 4
-2 -2
1 2

#include "stdio.h"#include "string.h"#include "math.h"#include "vector"using namespace std;struct node{int x,y;};vector<node> a[1005];bool change(int x1,int y1,int x2,int y2){bool flag = true;if     (-x1*x2+y1*y2==0 && y1!=y2) { printf("YES\n%d %d\n%d %d\n%d %d\n",0,0,-x1,y1,x2,y2); }else if(-x1*y2+y1*x2==0 && y1!=x2) { printf("YES\n%d %d\n%d %d\n%d %d\n",0,0,-x1,y1,y2,x2); }else if(-y1*x2+x1*y2==0 && x1!=y2) { printf("YES\n%d %d\n%d %d\n%d %d\n",0,0,-y1,x1,x2,y2); }else if(-y1*y2+x1*x2==0 && x1!=x2) { printf("YES\n%d %d\n%d %d\n%d %d\n",0,0,-y1,x1,y2,x2); }elseflag = false;return flag;}int main(){int i,j;int x,y;int ans,num;node cur;for(i=1; i<=1000; ++i){for(j=i; j<=1000; ++j){ans = (int)sqrt(i*i+j*j);if(ans > 1000) continue;if(ans*ans==i*i+j*j){cur.x = i;cur.y = j;a[ans].push_back(cur);}}}bool flag = false;scanf("%d %d",&x,&y);for(i=0; i<a[x].size(); i++){for(j=0; j<a[y].size(); j++){if( change(a[x][i].x,a[x][i].y,a[y][j].x,a[y][j].y) ){flag = true;break;}}if(flag) break;}if(!flag)printf("NO\n");return 0;}