[CodeForces 663E] - Binary Table(FWT)

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

mask对sta的影响，显然操作后的结果为mask ^ sta

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);
}