hdu 5605 /BC 68A geometry

设|PA|为z P点坐标(x,y) 则由相似三角形可得 |PB|=z*y*sqrt(z²-x²)。ans=|PA|*|PB|=z²*y*sqrt(z²-x²)

所以ans与y成正相关 因为 x y可互换 所以 ans也与x成正相关 由样例可知 ans=2*x*y

#include<iostream>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#include<vector>#include<queue>#include<map>#include<algorithm>#include<set>#define scnaf scanf#define cahr char#define bug puts("bugbugbug");using namespace std;typedef long long ll;const int mod=1000000007;const int maxn=150+5;const int inf=1e9;const double eps=1e-8;int main(){    int T_T;    scanf("%d",&T_T);    while(T_T--){        int x,y;        scanf("%d%d",&x,&y);        printf("%d\n",2*x*y);    }    return 0;}