http://codeforces.com/contest/838/problem/A

A. Binary Blocks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.

Input

The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.

The next n lines of input will contain a binary string with exactly m characters, representing the image.

Output

Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.

Example

input

3 5
00100
10110
11001

output

5

Note

We first choose k = 2.

The image is padded as follows:

001000
101100
110010
000000

We can toggle the image to look as follows:

001100
001100
000000
000000

We can see that this image is compressible for k = 2.

题意：给你一个n*m的0,1表，然后让你找到一个整数K去划分这张图，然后可以改变每个划分的图的任意的值，使每一部分的的值相同

题解：枚举k,维护0,1表的前缀和，去快速计算每个分块的总和；下面代码

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<iostream>
#define ll long long
using namespace std;
const int maxn=2e3+5e2+;
const int inf=0x3f3f3f3f;
int mp[maxn][maxn];
char a[maxn];
int n,m;
int main()
{
    scanf("%d %d",&n,&m);
    for(int i=;i<=n;i++)
    {
        scanf("%s",a+);
        for(int j=;j<=m;j++)
        {
            if(a[j]=='')
            mp[i][j]=;
        }
    }
    for(int i=;i<=n;i++)
    {
        for(int j=;j<=m;j++)
        {
            mp[i][j]=mp[i-][j]+mp[i][j-]+mp[i][j]-mp[i-][j-];
        }
    }
    int ans=inf;
    int len=max(n,m);
    for(int k=;k<=len;k++)
    {
        int nn=n,mm=m;
        if(n%k!=)nn=(n/k)*k+k;
        if(m%k!=)mm=(m/k)*k+k;
         int tmp1=;
        for(int i=;i<=nn/k;i++)
        for(int j=;j<=mm/k;j++)
        {
            int x1=i*k>n?n:i*k;
            int y1=j*k>m?m:j*k;
            int x2=i*k-k+>n?n:i*k-k+;
            int y2=j*k-k+>m?m:j*k-k+;
            int tmp=mp[x1][y1]+mp[x2-][y2-]-mp[x1][y2-]-mp[x2-][y1];
            if(tmp<=k*k/)
            {
                tmp1+=(tmp);
            }
            else
            {
                tmp1+=(k*k-tmp);
            }
        }
        ans=min(ans,tmp1);
    } 

    printf("%d\n",ans);
}