解题思路:
注意事项:
参考代码:
#include<stdio.h>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
int main(){
double x1,y1,x2,y2;
double x3,y3,x4,y4;
double m1,n1;
double m2,n2;
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
scanf("%lf%lf%lf%lf",&x3,&y3,&x4,&y4);
m1=max(min(x1,x2),min(x3,x4)); //左上角坐标,不管是第一种情况,还是第二种情况,最后都是正的
n1=max(min(y1,y2),min(y3,y4));
m2=min(max(x1,x2),max(x3,x4));
n2=min(max(y1,y2),max(y3,y4));
if(m2>m1 && n2>n1){
printf("%.2f\n",(m2-m1)*(n2-n1));
}else{
printf("0.00\n");
}
return 0;
}
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