codeforces C. Jzzhu and Chocolate
2015-03-14 16:37
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http://codeforces.com/contest/450/problem/C
题意:一个n×m的矩形,然后可以通过横着切竖着切,求切完k次之后最小矩形面积的最大值。
思路:设k1为横着切的次数,k2为竖着切的次数,最后的面积的大小为s=n/(k1+1)*(m/(k2+1)); 只有(k1+1)*(k2+1)的最小时,s最大.
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题意:一个n×m的矩形,然后可以通过横着切竖着切,求切完k次之后最小矩形面积的最大值。
思路:设k1为横着切的次数,k2为竖着切的次数,最后的面积的大小为s=n/(k1+1)*(m/(k2+1)); 只有(k1+1)*(k2+1)的最小时,s最大.
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #define ll long long using namespace std; ll n,m,k; int main() { cin>>n>>m>>k; ll ans=0; if(k>(n-1+m-1)) { printf("-1\n"); return 0; } else { if(n-1>=k) { ans=max(ans,m*(n/(k+1))); } else { ans=max(ans,m/(k-(n-1)+1)); } if(m-1>=k) { ans=max(ans,n*(m/(k+1))); } else { ans=max(ans,n/(k-m+2)); } } cout<<ans<<endl; return 0; }
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