题解 | #牛牛的线段#
牛牛的线段
https://www.nowcoder.com/practice/f72c56ed71664af082c921bf79861c85
#include<stdio.h>
#include<math.h>
int main()
{
int x1, y1 = 0;
int x2, y2 = 0;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int x = x1 - x2;
int y = y1 - y2;
int p = pow(x, 2) + pow(y, 2);
printf("%d", p);
return 0;
}
