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,y1,x2,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
int num=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
printf("%d\n",num);
return 0;
}
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