1 条题解
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1
#include <iomanip> #include <algorithm> using namespace std; int main() { double x1,y1,x2,y2; // 读第一个矩形 cin >> x1 >> y1 >> x2 >> y2; double r1_xmin = min(x1,x2); double r1_xmax = max(x1,x2); double r1_ymin = min(y1,y2); double r1_ymax = max(y1,y2); // 读第二个矩形 cin >> x1 >> y1 >> x2 >> y2; double r2_xmin = min(x1,x2); double r2_xmax = max(x1,x2); double r2_ymin = min(y1,y2); double r2_ymax = max(y1,y2); // 求交集边界 double ix1 = max(r1_xmin, r2_xmin); double iy1 = max(r1_ymin, r2_ymin); double ix2 = min(r1_xmax, r2_xmax); double iy2 = min(r1_ymax, r2_ymax); double area = 0.0; if(ix1 < ix2 && iy1 < iy2) { area = (ix2 - ix1) * (iy2 - iy1); } // 保留两位小数输出 cout << fixed << setprecision(2) << area << endl; return 0; }
- 1
信息
- ID
- 1050
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 10
- 标签
- 递交数
- 5
- 已通过
- 3
- 上传者