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, x2, y2;
scanf("%d %d", &x1, &y1);
getchar();
scanf("%d %d", &x2, &y2);
int ret = 0;
ret =(int)pow((x1 - x2), 2) + pow((y1 - y2), 2);
printf("%d\n", ret);
return 0;
}
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