题解 | 牛牛的线段

#include <stdio.h>
#include <math.h>
int main() {
    int x1,y1,x2,y2;
    int length;
    scanf("%d %d",&x1,&y1);
    scanf("%d %d",&x2,&y2);
    length = pow((x1 - x2),2) +pow((y1 - y2),2);
    printf("%d",length);
    return 0;
}