Codeforces #257 (Div. 2) C. Jzzhu and Chocolate
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简单数学题啊
就是抽象模型有点难,还有数据类型要注意!!
设横着划分为x行,竖着划分为y列
则最多可划分为(n/(x+1))*(m/(y+1))
注意一定要写成这种形式!!!原因在于n=3,m=3,x=1,y=1时这个式子和n*m/(x+1)/(y+1)的区别
其中x+y==k
所以分母可以划为开口向下一元二次方程
取其值范围的两个端点值,比较后输出最大值即可
注意要用long long不然会跪
代码如下:
#include <cstdio>#include <iostream>#include <algorithm>#define MAXN 10010#define LL long longusing namespace std;LL n, m, k;int main(void) { LL e1, e2, ans1, ans2, i; while(cin >> n >> m >> k) { if(n+m-2 < k) cout << "-1" << endl; else { e1 = max(k+1-m, 0ll); e2 = min(k, n-1); ans1 = e1; ans2 = e2; ans1 = max((n/(ans1+1))*(m/(k-ans1+1)), (n/(ans2+1))*(m/(k-ans2+1))); cout << ans1 << endl; } } return 0;}
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