Codeforces 450 C. Jzzhu and Chocolate
2014-07-20 00:03
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二次函数 n*m/ ( (x+1)*(k-x+1) ) 是对称轴>0开口向下的二次函数,直接考虑 x==0||x==n-1||x==m-1等进行判断即可.....
C. Jzzhu and Chocolate
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times.
Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical);
each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
![](http://espresso.codeforces.com/15ef03a74347ef850c56c54bb04b0a7b8fb3d1b3.png)
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants
this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece
is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 ≤ n, m ≤ 109; 1 ≤ k ≤ 2·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Sample test(s)
input
output
input
output
input
output
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
![](http://espresso.codeforces.com/8845791751b665bf2158530fb6f2b3e1ddc850d2.png)
In the second sample the optimal division looks like this:
![](http://espresso.codeforces.com/85864264c5cfc7d0dd783485087eeae6df30b934.png)
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long int LL;
LL n,m,k,ans=0LL;
int main()
{
cin>>n>>m>>k;
if(n+m-2<k)
{
puts("-1"); return 0;
}
if(n+m-2==k)
{
puts("1"); return 0;
}
LL t;
if(m>=k+1)
ans=max(ans,n*(m/(k+1)));
else
{
t=k-m+1;
ans=max(ans,n/(t+1));
}
if(n>=k+1)
ans=max(ans,m*(n/(k+1)));
else
{
t=k-n+1;
ans=max(ans,m/(t+1));
}
cout<<ans<<endl;
return 0;
}
C. Jzzhu and Chocolate
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times.
Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical);
each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
![](http://espresso.codeforces.com/15ef03a74347ef850c56c54bb04b0a7b8fb3d1b3.png)
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants
this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece
is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 ≤ n, m ≤ 109; 1 ≤ k ≤ 2·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Sample test(s)
input
3 4 1
output
6
input
6 4 2
output
8
input
2 3 4
output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
![](http://espresso.codeforces.com/8845791751b665bf2158530fb6f2b3e1ddc850d2.png)
In the second sample the optimal division looks like this:
![](http://espresso.codeforces.com/85864264c5cfc7d0dd783485087eeae6df30b934.png)
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long int LL;
LL n,m,k,ans=0LL;
int main()
{
cin>>n>>m>>k;
if(n+m-2<k)
{
puts("-1"); return 0;
}
if(n+m-2==k)
{
puts("1"); return 0;
}
LL t;
if(m>=k+1)
ans=max(ans,n*(m/(k+1)));
else
{
t=k-m+1;
ans=max(ans,n/(t+1));
}
if(n>=k+1)
ans=max(ans,m*(n/(k+1)));
else
{
t=k-n+1;
ans=max(ans,m/(t+1));
}
cout<<ans<<endl;
return 0;
}
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