题目

Codeforces 题目链接

分析

大佬博客,写的很好

本蒟蒻就不赘述了,就是一个看不出来的异或卷积

精髓在于

AC code

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;//必须用long long,过程中可能炸int
const int MAXN = 1<<20;
const LL INF = 1e18;
const int MAXM = 1e5 + 1;
int n, m, num[MAXM];
LL A[MAXN], B[MAXN];
char str[20][MAXM]; inline int calc(int s)
{
int ret = 0;
while(s) ++ret, s -= s&(-s);
return min(n - ret, ret);
} inline void FWT(LL arr[], const int &len, const int &flg)
{
register LL x, y;
for(register int i = 2; i <= len; i<<=1)
for(register int j = 0; j < len; j += i)
for(register int k = j; k < j + i/2; ++k)
{
x = arr[k], y = arr[k + i/2];
if(~flg) arr[k] = x + y, arr[k + i/2] = x - y;
else arr[k] = (x + y) / 2, arr[k + i/2] = (x - y) / 2;
}
} int main ()
{
scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i)
scanf("%s", str[i]);
for(int i = 0; i < m; ++i)
{
for(int j = 0; j < n; ++j)
num[i] |= (str[j][i] - '0') << j;
++A[num[i]];
}
for(int i = 0; i < (1<<n); ++i)
B[i] = calc(i);
FWT(A, 1<<n, 1);
FWT(B, 1<<n, 1);
for(int i = 0; i < (1<<n); ++i)
A[i] *= B[i];
FWT(A, 1<<n, -1);
LL Ans = INF;
for(int i = 0; i < (1<<n); ++i)
Ans = min(Ans, A[i]);
printf("%I64d\n", Ans);
}
05-11 17:54