[ABC223E] Placing Rectangles 题解

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[ABC223E] Placing Rectangles 题解

思路解析

根据题目可知，其实三个长方形无非只有以下两种摆放方式。

若大长方形长为 $y$，宽为 $x$，则我们对于第一种情况就固定住宽，判断能否使长度小于等于长；对于第二种情况同样固定住宽，此时 A 长方形右边空间的长就确定了，就只需要判断 B,C 的宽之和能否小于大长方形的宽即可。

注意大长方形的长宽可以互换，小长方形的顺序可以互换。

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define double long double
ll x, y, a, b, c;
bool check() {
	ll ax = ceil((double)a / x), bx = ceil((double)b / x), cx = ceil((double)c / x);
	if(ax + bx + cx <= y) return true;
	ll yy = y - ax;
	if(yy <= 0) return false;
	ll by = ceil((double)b / yy), cy = ceil((double)c / yy);
	if(by + cy <= x) return true;
	return false;
}
int main() {
	cin >> x >> y >> a >> b >> c;
	bool ans = check();
	swap(a, b); ans |= check();
	swap(a, c); ans |= check();
	
	swap(x, y); ans |= check();
	swap(a, b); ans |= check();
	swap(a, c); ans |= check();
	if(ans) puts("Yes");
	else puts("No");
	return 0;
}