CodeForces 449 A.Jzzhu and Chocolate(枚举)
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Description
一个n*m的网格,在网格线上横纵共切k刀,使得分成小块的面积最小值最大化
Input
三个整数n,m,k(1<=n,m<=1e9,1<=k<=2e9)
Output
输出最大化的面积最小值
Sample Input
3 4 1
Sample Output
6
Solution
设横着x刀,则纵着k-x刀,为使最小值最大化显然要尽量均匀切,那么这样切得到的面积最小值应该为
由于取值至多种,所以枚举的值求出对应的最小面积并从中取最小值即为答案
Code
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<map>#include<set>#include<ctime>using namespace std;typedef long long ll;ll Solve(ll n,ll m,ll k){ ll ans=1; for(int i=1,pre=0;i<=n;i=pre+1) { pre=n/(n/i); if(pre>k+1) { ans=max(ans,n/(k+1)*m); break; } ans=max(ans,(n/pre)*(m/(k-pre+2))); } return ans;}int main(){ ll n,m,k; while(~scanf("%I64d%I64d%I64d",&n,&m,&k)) { if(k>n+m-2)printf("-1\n"); else printf("%I64d\n",Solve(n,m,k)); } return 0;}
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