LOJ #2463. 「2018 集训队互测 Day 1」完美的旅行（BM）

题目
枚举与值，枚举步数和最后一个点DP，枚举前一个点转移，只把起点都终点符合与值的方案计算。
最后得到的是与值为其超集的方案数，容斥一下（你也可以叫这IFMT）就可以得出答案。
时间复杂度$O(n^{3}m)$
循环展开+使用$\text{unsigned long long}$代替$\text{long long}$ 获得$\text{1000ms}$加速即可AC此题。
循环展开太强辣

AC Code:

#include 
#define maxn 64
#define maxm 20005
#define mod 998244353
#define LL unsigned long long
using namespace std;

int n, m, A[maxn][maxn], B[maxn][maxn], S[maxn], f[2][maxn], ans[maxn][maxm];
inline int add(int a, int b) {
     
    int c = a + b;
    return c >= mod ? c - mod : c;
}

int main() {
     
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) scanf("%d", &A[i][j]);
    for (int x = 0; x < n; x++) {
     
        memset(f, 0, sizeof f);
        memset(S, 0, sizeof S);
        int now = 1, pre = 0;
        for (int i = 0; i < n; i++) {
     
            if ((i & x) == x) {
     
                for (int j = 0; j < n; j++) S[j] = add(S[j], A[i][j]);
                f[pre][i] = 1;
            }
        }
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++) B[i][j] = add(A[i][j], ((i & x) == x) * S[j]);
        for (int i = 0; i < n; i++) {
     
            LL ret = 0;
            for (int k = 0; k < n; k += 8) {
     
                ret += (LL)f[pre][k] * A[k][i], ret += (LL)f[pre][k + 1] * A[k + 1][i],
                    ret += (LL)f[pre][k + 2] * A[k + 2][i], ret += (LL)f[pre][k + 3] * A[k + 3][i],
                    ret += (LL)f[pre][k + 4] * A[k + 4][i], ret += (LL)f[pre][k + 5] * A[k + 5][i],
                    ret += (LL)f[pre][k + 6] * A[k + 6][i], (ret += (LL)f[pre][k + 7] * A[k + 7][i]) %= mod;
            }
            f[now][i] = ret;
            if ((i & x) == x)
                ans[x][1] = add(ans[x][1], ret);
        }
        swap(now, pre);
        for (int j = 2; j <= m; j++, swap(now, pre))
            for (int i = 0; i < n; i++) {
     
                LL ret = 0;
                for (int k = 0; k < n; k += 8) {
     
                    ret += (LL)f[pre][k] * B[k][i], ret += (LL)f[pre][k + 1] * B[k + 1][i],
                        ret += (LL)f[pre][k + 2] * B[k + 2][i], ret += (LL)f[pre][k + 3] * B[k + 3][i],
                        ret += (LL)f[pre][k + 4] * B[k + 4][i], ret += (LL)f[pre][k + 5] * B[k + 5][i],
                        ret += (LL)f[pre][k + 6] * B[k + 6][i],
                        (ret += (LL)f[pre][k + 7] * B[k + 7][i]) %= mod;
                }
                f[now][i] = ret;
                if ((i & x) == x)
                    ans[x][j] = add(ans[x][j], ret);
            }
    }
    for (int bit = 1; bit < n; bit <<= 1)
        for (int i = 0; i < n; i++)
            if (i & bit)
                for (int j = 1; j <= m; j++) ans[i - bit][j] = add(ans[i - bit][j], mod - ans[i][j]);
    int res = 0;
    for (int i = 0; i < n; i++)
        for (int j = 1; j <= m; j++) res ^= ans[i][j];
    printf("%d\n", res);
}

正解：就是上面的$dp$只不过我们对于步数只$dp$到$2n+1$步，然后对于每个值拿$a[0...2n+1]$做$BM$然后就可以$O(n^{2}m)$求出与值为其超集的方案数。
注意是一次旅行是$n$阶递推，也就是说一次旅行是可以写作$\frac{S(x)}{R(x)}$的形式，其中$R(x)$的次数$\le n$,$S(x)$的次数$\le n-1$，那么多次旅行就是$\frac{1}{1-\frac{S(x)}{R(x)}}=\frac{R(x)}{R(x)-S(x)}$，所以分子的次数是可以达到$n$次的，所以多次旅行是$n+1$阶递推，需要前$2n+1$项（虽然实际测试$2n$项就可以过，感觉是因为最高次是分子的原因。）

$Code$

#include
#define maxn 65
#define maxm 20005
#define rep(i,j,k) for(int i=(j),LIM=(k);i<=LIM;i++)
#define per(i,j,k) for(int i=(j),LIM=(k);j>=LIM;j--)
#define mod 998244353
#define vi vector
#define pb push_back
#define poly vi
using namespace std;

int S;
int n,L,m,a[maxn][maxn],b[maxn][maxn],c[maxn][maxn],g[maxn<<1][maxn],f[maxm][maxn];
int Pow(int b,int k){
      int r=1;for(;k;k>>=1,b=1ll*b*b%mod) if(k&1) r=1ll*r*b%mod;return r; }
void FMT(int *a,int tp){
     
	rep(j,0,L-1) rep(i,0,n-1)
		if(i >> j & 1){
     
			int v = (i ^ (1 << j));
			if(tp == 1) a[v] = (a[v] + a[i]) % mod;
			else a[v] = (a[v] - a[i]) % mod;
		}
}
vi BM(vi &a){
     
	vi r(1,1),p,t;int l;
	rep(i,0,a.size()-1){
     
		int b=0;
		rep(j,0,r.size()-1) b = (b + 1ll * r[j] * a[i-j]) % mod;
		if(!b) continue;
		t = r , r.resize(max(r.size() , i+2 - (r.size() - (r.size() != 0))));
		rep(j,0,p.size()-1) r[i-l+j] = (r[i-l+j] - 1ll * p[j] * b) % mod;
		p = t;int iv = Pow(b , mod-2);
		rep(j,0,p.size()-1) p[j] = 1ll * p[j] * iv % mod;
		l = i;
	}
	return r;
}

int main(){
     
	scanf("%d%d",&n,&m);
	for(L=0;(1<<L) < n;L++);
	rep(i,0,n-1) rep(j,0,n-1) scanf("%d",&a[i][j]);
	rep(i,0,n-1) b[i][i] = 1;
	rep(i,0,n-1) f[0][i] = 1;
	int S = n * 2 + 1;
 	rep(i,1,S){
     
		rep(j,0,n-1) rep(k,0,n-1) rep(p,0,n-1) c[j][p] = (c[j][p] + 1ll * a[j][k] * b[k][p]) % mod;
		rep(j,0,n-1) rep(k,0,n-1) g[i][j&k] = (g[i][j&k] + (b[j][k] = c[j][k])) % mod , c[j][k] = 0;
		FMT(g[i],1);
	}
	rep(i,1,S) rep(j,1,i) rep(k,0,n-1) f[i][k] = (f[i][k] + 1ll * f[i-j][k] * g[j][k]) % mod; 
	rep(i,0,n-1){
     
		vi a;
		rep(j,0,S) a.pb(f[j][i]);
		vi P = BM(a);
		rep(j,S+1,m){
     
			f[j][i] = 0;
			rep(k,1,P.size()-1)
				f[j][i] = (f[j][i] - 1ll * f[j-k][i] * P[k]) % mod;
		}
	}
	int ans = 0;
	rep(i,1,m) FMT(f[i],-1);
	rep(i,1,m) rep(j,0,n-1) ans ^= (f[i][j] + mod) % mod;
	printf("%d\n",ans);
}