题解 | #牛牛的线段#
牛牛的线段
https://www.nowcoder.com/practice/f72c56ed71664af082c921bf79861c85
#include <stdio.h> int main() { int a,b,c,d; scanf("%d %d %d %d",&a,&b,&c,&d); printf("%d",(a-c)*(a-c)+(b-d)*(b-d)); return 0; }
牛牛的线段
https://www.nowcoder.com/practice/f72c56ed71664af082c921bf79861c85
#include <stdio.h> int main() { int a,b,c,d; scanf("%d %d %d %d",&a,&b,&c,&d); printf("%d",(a-c)*(a-c)+(b-d)*(b-d)); return 0; }
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