LOJ #6503. 「雅礼集训 2018 Day4」Magic

多项式复习好题

题意

有$m$种小球,每种小球有$a[i]$个,一共有$\sum a[i]=n$个小球
把$n$个小球排成一排
问恰好有$k$对相邻的小球种类相同的方案数

题解

对于64’

直接做恰好k对的不好做,我们可以先做出至少k对的来
$设f[i]表示至少i对的方案数,g[i]为恰好i对的方案数$
$可以发现g[i]=f[i]-\sum _{j=i+1}^{n}g[j]\ast C_j^i$
$f[i]=\sum _{j=i}^n{C}_j^i\ast g[j]$
$根据二项式反演g[i]=\sum _{j=i}^n(-1{)}^{j-i}\ast C_j^{i}f[j]$

如何求出至少k对???
相邻的一对种类相同的球可以把这一对合成一个球
考虑DP
$设dp[i][j]表示前i种颜色,分成j块(相邻两块不一定种类不同)$
$可得dp[i][j]=\sum dp[i-1][j-k]\ast C_{a[i]-1}^{k-1}\ast C_j^k$

意思就是把当前种类的球分成$k$份,就是$\ast C_{a[i]-1}^{k-1}$在插入到块中就是$\ast C_j^k$

$最后f[i]=dp[m][n-i],再通过二项式反演求出g[i]即可$

这样就可以获得64’的好成绩

code:

#include<bits/stdc++.h>
#define mod 998244353
#define ll long long
using namespace std;
int n, a[100005];
ll f[2][100005], fac[100005], _fac[100005], g[100005];
ll qpow(ll x, ll y) {
	ll ret = 1;
	for(; y; y >>= 1, x = x * x % mod) if(y & 1) ret = ret * x % mod;
	return ret;
}
ll C(int x, int y) {
	if(x < y) return 0;
	return fac[x] * _fac[x - y] % mod * _fac[y] % mod;
}//组合数计算 
int main() {
	fac[0] = 1;
	for(int i = 1; i <= 100000; i ++) fac[i] = fac[i - 1] * i % mod;
	_fac[100000] = qpow(fac[100000], mod - 2);
	for(int i = 100000 - 1; i >= 0; i --) _fac[i] = _fac[i + 1] * (i + 1) % mod;//预处理fac和invfac用于算组合数 
	int q, Q;
	scanf("%d%d%d", &n, &Q, &q);//这里的n,m和题目给出的是不同的 
	for(int i = 1; i <= n; i ++) scanf("%d", &a[i]);
	int m = a[1];
	for(int i = 1; i <= m; i ++) f[1][i] = C(a[1] - 1, i - 1);
	
	for(int i = 2; i <= n; i ++) {
		memset(f[i&1], 0, sizeof f[i&1]);
		m += a[i];
		for(int j = 1; j <= m; j ++) {
			for(int k = 1; k <= min(j, a[i]); k ++){
				f[i&1][j] += f[(i - 1)&1][j - k] * C(a[i] - 1, k - 1) % mod * C(j, k) % mod, f[i&1][j] %= mod;//DP 
			}
		}
	} 
	for(int i = 0; i <= m; i ++) g[i] = f[n&1][m - i];
		ll ans = 0;
		for(int i = q; i <= m; i ++) {//二项式反演 
		if((i - q) & 1) ans -= C(i, q) * g[i] % mod, ans = (ans + mod) % mod; 
		else ans += C(i, q) * g[i] % mod, ans %= mod;
	}
	printf("%lld\n", ans);//输出 

	return 0;
}

对于100’

我们再看一下转移的这条式子

$dp[i][j]=\sum dp[i-1][j-k]\ast C_{a[i]-1}^{k-1}\ast C_j^k$

设$dp[i][j]为f[j],dp[i-1][j-k]为g[j-k]$
可得

$f[j]=\sum g[j-k]\ast C_{a[i]-1}^{k-1}\ast C_j^k$

按照一般的讨论都是要把组合数拆开
肯定先拆$C_j^k$因为它和$j,k$都有关

$f[j]=\sum g[j-k]\ast C_{a[i]-1}^{k-1}\ast \frac{j!}{(j-k)!k!}$

把$j!$除过去

$\frac{1}{j!}f[j]=\sum \frac{1}{(j-k)!}g[j-k]\ast C_{a[i]-1}^{k-1}\ast \frac{1}{k!}$

$设ff[j]=\frac{1}{j!}f[j],gg[i][k]=C_{a[i]-1}^{k-1}\ast \frac{1}{k!}$

$可得ff[j]=\sum ff[j-k]\ast gg[i][k]$

这明显就是一个卷积的形式(分治FFT)

没有一眼看出来的可以考虑生成函数

$记F_i=\sum _{j=1}^{\infty }{C}_{a[i]-1}^{j-1}\frac{x^j}{j!}$

$明显答案就是ans=\prod F_i$

这个东西也是直接分治+NTT就可以了
大概是O(nlog^2n)
code:

#include<bits/stdc++.h>
#define N 400005
#define mod 998244353
#define G 3 
using namespace std;
inline int add(int x, int y) {
	x += y; return (x >= mod)? x - mod : x;
}
inline int sub(int x, int y) {
	x -= y; return (x < 0)? x + mod : x;
}
inline int mul(int x, int y) {
	return 1ll * x * y % mod;
}
inline int qpow(int x, int y) {
	if(y < 0) {y = -y; x = qpow(x, mod - 2);}
	int ans = 1;
	for(; y; y >>= 1, x = mul(x, x)) if(y & 1) ans = mul(ans, x);
	return ans;
}
int fac[N], invfac[N], rev[N];
void init() {
	fac[0] = 1;
	for(int i = 1; i < N; i ++) fac[i] = mul(fac[i - 1], i);
	invfac[N - 1] = qpow(fac[N - 1], mod - 2);
	for(int i = N - 2; i >= 0; i --) invfac[i] = mul(invfac[i + 1], i + 1);
}
inline int C(int x, int y) {
	if(x < y || y < 0) return 0;
	return mul(fac[x], mul(invfac[x - y], invfac[y]));
}
void ntt(int *a, int len, int o) {
	for(int i = 1; i < len; i ++) rev[i] = (rev[i >> 1] >> 1) | (((i & 1) * len) >> 1);
	for(int i = 1; i < len; i ++) if(i < rev[i]) swap(a[i], a[rev[i]]);
	for(int i = 1; i < len; i <<= 1) {
		int wn = qpow(G, o * (mod - 1) / (i << 1));
		for(int j = 0; j < len; j += i << 1) {
			int wnk = 1;
			for(int k = j; k < j + i; k ++, wnk = mul(wnk, wn)) {
				int X = a[k], Y = mul(a[k + i], wnk);
				a[k] = add(X, Y);
				a[k + i] = sub(X, Y);
			}
		}
	}
	if(o == -1) {
		int invlen = qpow(len, mod - 2);
		for(int i = 0; i < len; i ++) a[i] = mul(a[i], invlen);
	}
}
int fa[N], ga[N];
void mult(vector<int> &ff, vector<int> &gg, vector<int> &fg) {
	for(int i = 0; i < ff.size(); i ++) fa[i] = ff[i];
	for(int i = 0; i < gg.size(); i ++) ga[i] = gg[i];
	int len = 1, l = ff.size() + gg.size() - 1;
	for(;len < l; len <<= 1);
	
	for(int i = ff.size(); i < len; i ++) fa[i] = 0;
	for(int i = gg.size(); i < len; i ++) ga[i] = 0;
	
	ntt(fa, len, 1), ntt(ga, len, 1);
	for(int i = 0; i < len; i ++) fa[i] = mul(fa[i], ga[i]);
	ntt(fa, len, -1);
	
	for(int i = 0; i < l; i ++) fg.push_back(fa[i]);
}//多项式乘法板子 
vector<int> ff[N], f[N];
void solve(int l, int r, int rt) {//分治+NTT 
	if(l == r) {
		ff[rt] = f[l];
		return;
	}
	int mid = (l + r) >> 1;
	solve(l, mid, rt << 1);
	solve(mid + 1, r, rt << 1 | 1);
	mult(ff[rt << 1], ff[rt << 1 | 1], ff[rt]);
}
int n, m, Q, q, g[N], a[N];
int main() {
	init();
	scanf("%d%d%d", &n, &Q, &q);
	for(int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]); m += a[i];
		f[i].push_back(0);
		for(int j = 1; j <= a[i]; j ++) f[i].push_back(mul(C(a[i] - 1, j - 1), invfac[j]));
	}
	solve(1, n, 1);
	for(int i = 0; i <= m; i ++) g[i] = mul(ff[1][m - i], fac[m - i]);
	int ans = 0;
	for(int i = q; i <= m; i ++) {//二项式反演 
		if((i - q) & 1) ans = sub(ans, mul(C(i, q), g[i]));
		else ans = add(ans, mul(C(i, q), g[i]));
	}
	printf("%d\n", ans);
	return 0;
}

考虑这题的加强版,多次询问k

因为每次求k的时候都要二项式反演一遍,所以时间复杂度会变成O(nq)

让我们再看一下二项式反演的那条式子

$g[i]=\sum _{j=i}^n(-1{)}^{j-i}\ast C_j^{i}f[j]$

把$g,f$反过来一下,方便看

$f[i]=\sum _{j=i}^n(-1{)}^{j-i}\ast C_j^{i}g[j]$

按照套路,先把组合数拆开

$f[i]=\sum _{j=i}^n(-1{)}^{j-i}\ast \frac{j!}{(j-i)!i!}g[j]$

把i!乘过去

$i!f[i]=\sum _{j=i}^n\frac{(-1{)}^{j-i}}{(j-i)!}\ast j!g[j]$

再按照套路,将g反转
即设$g^{\prime }[j]=g[n-j]$

$i!f[i]=\sum _{n-j=i}^n\frac{(-1{)}^{n-j-i}}{(n-j-i)!}\ast (n-j)!g^{\prime }[j]$

$i!f[i]=\sum _{j=n-i}^n\frac{(-1{)}^{n-j-i}}{(n-j-i)!}\ast (n-j)!g^{\prime }[j]$

再把中间那坨反转一下(即n-i=i)

$(n-i)!f[n-i]=\sum _{j=0}^i\frac{(-1{)}^{i-j}}{(i-j)!}\ast (n-j)!g^{\prime }[j]$

再按照套路,将f反转
即设$f^{\prime }[i]=f[n-i]$

$(n-i)!f^{\prime }[i]=\sum _{j=0}^i\frac{(-1{)}^{i-j}}{(i-j)!}\ast (n-j)!g^{\prime }[j]$

设$h[i]=(n-i)!f^{\prime }[i],ff[i]=\frac{(-1{)}^i}{i!},gg[i]=(n-i)!g^{\prime }[i]$

$然后发现h=ff\ast gg$

xjb卷一下就行了QWQ
code:

#include<bits/stdc++.h>
#define N 4000005
#define mod 998244353
#define G 3 
using namespace std;
inline int add(int x, int y) {
	x += y; return (x >= mod)? x - mod : x;
}
inline int sub(int x, int y) {
	x -= y; return (x < 0)? x + mod : x;
}
inline int mul(int x, int y) {
	return 1ll * x * y % mod;
}
inline int qpow(int x, int y) {
	if(y < 0) {y = -y; x = qpow(x, mod - 2);}
	int ans = 1;
	for(; y; y >>= 1, x = mul(x, x)) if(y & 1) ans = mul(ans, x);
	return ans;
}
int fac[N], invfac[N], rev[N];
void init() {
	fac[0] = 1;
	for(int i = 1; i < N; i ++) fac[i] = mul(fac[i - 1], i);
	invfac[N - 1] = qpow(fac[N - 1], mod - 2);
	for(int i = N - 2; i >= 0; i --) invfac[i] = mul(invfac[i + 1], i + 1);
}
inline int C(int x, int y) {
	if(x < y || y < 0) return 0;
	return mul(fac[x], mul(invfac[x - y], invfac[y]));
}
void ntt(int *a, int len, int o) {
	for(int i = 1; i < len; i ++) rev[i] = (rev[i >> 1] >> 1) | (((i & 1) * len) >> 1);
	for(int i = 1; i < len; i ++) if(i < rev[i]) swap(a[i], a[rev[i]]);
	for(int i = 1; i < len; i <<= 1) {
		int wn = qpow(G, o * (mod - 1) / (i << 1));
		for(int j = 0; j < len; j += i << 1) {
			int wnk = 1;
			for(int k = j; k < j + i; k ++, wnk = mul(wnk, wn)) {
				int X = a[k], Y = mul(a[k + i], wnk);
				a[k] = add(X, Y);
				a[k + i] = sub(X, Y);
			}
		}
	}
	if(o == -1) {
		int invlen = qpow(len, mod - 2);
		for(int i = 0; i < len; i ++) a[i] = mul(a[i], invlen);
	}
}
int fa[N], ga[N];
void mult(vector<int> &ff, vector<int> &gg, vector<int> &fg) {
	for(int i = 0; i < ff.size(); i ++) fa[i] = ff[i];
	for(int i = 0; i < gg.size(); i ++) ga[i] = gg[i];
	int len = 1, l = ff.size() + gg.size() - 1;
	for(;len < l; len <<= 1);
	
	for(int i = ff.size(); i < len; i ++) fa[i] = 0;
	for(int i = gg.size(); i < len; i ++) ga[i] = 0;
	
	ntt(fa, len, 1), ntt(ga, len, 1);
	for(int i = 0; i < len; i ++) fa[i] = mul(fa[i], ga[i]);
	ntt(fa, len, -1);
	
	for(int i = 0; i < l; i ++) fg.push_back(fa[i]);
}
vector<int> ff[N], f[N];
void solve(int l, int r, int rt) {
	if(l == r) {
		ff[rt] = f[l];
		return;
	}
	int mid = (l + r) >> 1;
	solve(l, mid, rt << 1);
	solve(mid + 1, r, rt << 1 | 1);
	mult(ff[rt << 1], ff[rt << 1 | 1], ff[rt]);
}
int n, m, Q, q, g[N], a[N];
int main() {
	init();
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]); m += a[i];
		f[i].push_back(0);
		for(int j = 1; j <= a[i]; j ++) f[i].push_back(mul(C(a[i] - 1, j - 1), invfac[j]));
	}
	solve(1, n, 1);
	for(int i = 0; i <= m; i ++) g[i] = mul(ff[1][m - i], fac[m - i]);
	//--------以下部分求h-------- 
	memset(fa, 0, sizeof fa);
	memset(ga, 0, sizeof ga);
	for(int i = 0; i <= m; i ++) {
		if(i & 1) fa[i] = sub(0, invfac[i]);
		else fa[i] = invfac[i];
	} 
	for(int i = 0; i <= m; i ++) {
		ga[i] = mul(fac[m - i], g[m - i]);
	}
	
	int len = 1;
	for(;len < m + m + 1;) len <<= 1; 
	ntt(fa, len, 1), ntt(ga, len, 1);
	for(int i = 0; i < len; i ++) fa[i] = mul(fa[i], ga[i]);
	ntt(fa, len, -1);//xjb卷起来 
	
	scanf("%d", &Q);
	while(Q --) {	
		scanf("%d", &q);
		printf("%d\n", mul(invfac[q], fa[m - q]));//记得乘1/q! 
	}
	
	return 0;
}

emmmmmm…多项式的香气
感觉越来越喜欢多项式了
这真是道有意思的题QWQ