题解 | #牛牛的线段#
牛牛的线段
http://www.nowcoder.com/practice/f72c56ed71664af082c921bf79861c85
int main(){
int x1 = 0;
int x2 = 0;
int y1 = 0;
int y2 = 0;
scanf("%d %d",&x1,&y1);
scanf("%d %d",&x2,&y2);
printf("%d",(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
return 0;
}
