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