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