题解 | 牛牛的线段

#include <stdio.h>

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