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