註意到n很小，顯然的做法是枚舉每行是否翻轉，然後O(m)統計Σmin(popcount(a_i),n-popcount(a_i))即可。考慮將每列是否翻轉寫成一個二進制數，那麽翻轉相當於讓該二進制數與每列異或。統計每列各種狀態的出現次數，將其設為cnt[i]，將min(popcount(i),n-popcount(i))設為g[i]，可以發現若翻轉狀態為j，答案即為Σcnt[i]·g[i^j]。這就是一個異或卷積，FWT就行了。

#include<bits/stdc++.h>
using namespace std;
int getbit(){char c=getchar();while
 
 (c<‘0‘||c>‘9‘) c=getchar();return c^48;}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f; 
}
#define N (1<<20)
#define M 100010
#define
 
 P 1000000007 
#define inv2 500000004
int n,m,a[M],cnt[N],g[N],ans;
void FWT(int *a,int n,int op)
{
    for (int i=2;i<=n;i<<=1)
        for (int j=0;j<n;j+=i)
            for (int k=j;k<j+(i>>1);k++)
            {
                int x=a[k],y=a[k+(i>>1)];
                a[k]=(x+y)%P,a[k+(i>>1
 
)]=(x-y+P)%P;
                if (op) a[k]=1ll*a[k]*inv2%P,a[k+(i>>1)]=1ll*a[k+(i>>1)]*inv2%P;
            }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=0;i<n;i++)
        for (int j=1;j<=m;j++)
        a[j]|=getbit()<<i;
    for (int i=1;i<=m;i++) cnt[a[i]]++;
    for (int i=1;i<(1<<n);i++) g[i]=g[i^(i&-i)]+1;
    for (int i=1;i<(1<<n);i++) g[i]=min(g[i],n-g[i]);
    FWT(g,(1<<n),0),FWT(cnt,(1<<n),0);
    for (int i=0;i<(1<<n);i++) g[i]=1ll*g[i]*cnt[i]%P;
    FWT(g,(1<<n),1);
    ans=n*m;
    for (int i=0;i<(1<<n);i++) ans=min(ans,g[i]);
    cout<<ans;
    return 0;
}

CF662C Binary Table（FWT）