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  • 【Bzoj4555】【Luogu P4091】求和(NTT)

    题面

    Bzoj

    Luogu

    题解

    先来颓柿子

    $$ sum_{i=0}^nsum_{j=0}^iS(i,j)2^jj! \ =sum_{j=0}^n2^jj!sum_{i=0}^nS(i,j) \ ecause S(n, m)=frac1{m!}sum_{i=0}^m(-1)^iinom{m}{i}(m-i)^n=sum_{i=0}^mfrac{(-1)^i}{i!}frac{(m-i)^n}{(m-i)!} \ herefore=sum_{j=0}^n2^jj!sum_{i=0}^nsum_{k=0}^{j}frac{(-1)^k}{k!}frac{(j-k)^i}{(j-k)!} \ =sum_{j=0}^n2^jj!sum_{k=0}^jfrac{(-1)^k}{k!}frac{sum_{i=0}^n(j-k)^i}{(j-k)!} \ =sum_{j=0}^n2^jj!sum_{k=0}^jfrac{(-1)^k}{k!}frac{(j-k)^{n+1}-1}{(j-k-1)(j-k)!} $$

    然后后面那一大坨可以看做卷积,因为要取模,$NTT$就好了。

    #include <cstdio>
    #include <algorithm>
    using std::swap;
    
    const int N = 2.7e5 + 10, Mod = 998244353, g = 3;
    int n, m, P, jc[N], pow2[N], invjc[N];
    int a[N], b[N], r[N], ret;
    
    int qpow(int a, int b) {
    	int ret = 1;
    	while(b) {
    		if(b & 1) ret = 1ll * ret * a % Mod;
    		a = 1ll * a * a % Mod, b >>= 1;
    	} return ret;
    }
    
    void NTT (int f[], int opt) {
    	for(int i = 0; i < n; ++i) if(i < r[i]) swap(f[i], f[r[i]]);
    	for(int len = 1, nl = 2; len < n; len = nl, nl <<= 1) {
    		int rot = qpow(g, (Mod - 1) / nl);
    		if(opt == -1) rot = qpow(rot, Mod - 2);
    		for(int l = 0; l < n; l += nl) {
    			int w = 1, r = l + len;
    			for(int k = l; k < r; ++k, w = 1ll * w * rot % Mod) {
    				int x = f[k], y = 1ll * f[k + len] * w % Mod;
    				f[k] = (x + y) % Mod, f[k + len] = (x + Mod - y) % Mod;
    			}
    		}
    	}
    }
    
    int main () {
    	scanf("%d", &n), jc[0] = pow2[0] = invjc[0] = b[0] = 1, b[1] = n + 1;
    	for(int i = 1; i <= n; ++i)
    		jc[i] = 1ll * jc[i - 1] * i % Mod, pow2[i] = (pow2[i - 1] << 1) % Mod;
    	invjc[n] = qpow(jc[n], Mod - 2);
    	for(int i = n - 1; i; --i) invjc[i] = 1ll * invjc[i + 1] * (i + 1) % Mod;
    	for(int i = 0; i <= n; ++i) a[i] = 1ll * invjc[i] * (i & 1 ? Mod - 1 : 1) % Mod;
    	for(int i = 2; i <= n; ++i)
    		b[i] = 1ll * (qpow(i, n + 1) + Mod - 1) % Mod * qpow(i - 1, Mod - 2) % Mod * invjc[i] % Mod;
    	for(m = n << 1, n = 1; n <= m; n <<= 1, ++P);
    	for(int i = 0; i < n; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (P - 1));
    	NTT(a, 1), NTT(b, 1);
    	for(int i = 0; i < n; ++i) a[i] = 1ll * a[i] * b[i] % Mod;
    	NTT(a, -1); int invn = qpow(n, Mod - 2);
    	for(int i = 0; i <= n; ++i)
    		ret = (ret + 1ll * pow2[i] * jc[i] % Mod * a[i] % Mod * invn % Mod) % Mod;
    	printf("%d
    ", ret);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/water-mi/p/10198928.html
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