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