hdu 6116 路径计数 百度之星初赛B补--计数问题+NTT

Posted 2020-10-10 HITLJR

题意和这个例题很像。交错排列问题。直接用三次NTT优化

 

#include <bits/stdc++.h>
const long long MOD = 998244353;
const double ex = 1e-10;
typedef long long LL;
#define inf 0x3f3f3f3f
using namespace std;
const int N = 16064;
const int p = 998244353;
const int G = 3;
const int NUM = 25;
LL x1[N],x2[N],wn[NUM];
LL F[N],Finv[N],inv[N];
inline LL quick_mod(LL a,LL b,LL m){
    LL ans = 1;
    a %= m;
    while (b){
        if (b % 2 == 1)ans =  ans * a % m;
        b/=2;
        a = a * a % m;
    }
    return ans;
}
inline void GetWn(){
    for (int i = 0 ; i <NUM ; i++){
        int t = 1 << i;
        wn[i] = quick_mod(G,(p-1)/t,p);
    }
}
inline void Rader(LL a[],int len){
    int j = len >> 1;
    for (int i = 1 ; i< len- 1; i++){
        if (i <j) swap(a[i],a[j]);
        int k =  len >> 1;
        while( j >= k){
            j-=k;k>>=1;
        }
        if ( j < k ) j+=k;
    }
}
inline void NTT(LL a[],int len,int on){
    Rader(a,len);
    int id = 0;
    for (int h = 2; h <=len ; h <<=1){
        id ++;
        for (int j = 0 ; j< len ; j+=h){
            LL w = 1;
            for (int k = j ; k < j + h/2 ; k++){
                LL u = a[k] % p;
                LL t = w * ( ( a[k+h/2] ) % p ) % p; // 注意a的下标
                a[k] = (u+t) %p;
                a[k + h / 2] = (((u-t) % p) + p) % p;
                w = w * wn[id] % p;
            }
        }
    }
    if (on == -1){
        for (int i = 1; i<len/2 ; i++){
            swap(a[i],a[len-i]);
        }
        LL Inv = quick_mod(len,p-2,p);
        for (int i = 0; i<len; i++){
            a[i] = a[i] % p * Inv % p;
        }
    }
}
inline void  conv(LL a[],LL b[],int n){
    NTT(a,n,1);
    NTT(b,n,1);
    for (int i = 0 ; i < n; i++)
        a[i] = a[i] * b[i] % p;
    NTT(a,n,-1);
}
inline void init(){
    inv[1] = 1;
    for (int i = 2; i<N; i++){
        inv[i] = (MOD-MOD/i) *1ll *inv[MOD % i] % MOD;
    }
    F[0] = Finv[0] = 1;
    for (int i = 1 ;i<N; i++){
        F[i] = F[i-1] * i % MOD;
        Finv[i] = Finv[i-1] *1ll*inv[i] % MOD;
    }
    return;
}
LL A[5][16064];
inline void getA(int d,int id){
    for (int i = 1 ; i <= d ; i++){
        A[id][i] = F[d-1]*Finv[i-1]%MOD * Finv[d-i]  % MOD * Finv[i] % MOD;
    }
}
int main()
{
    init();
    GetWn();
    int aa[5];
    while (cin >> aa[1] >> aa[2] >> aa[3] >> aa[4]){
        memset(A,0,sizeof(A));
        int len = 0;
        for (int i = 1; i<=4; i++){
            getA(aa[i],i);
            len += aa[i];
        }
        int l = 1;
        while (l < 2 * (len + 1)) l<<=1; // l 为扩展长度
        conv(A[1],A[2],l);
        conv(A[1],A[3],l);
        conv(A[1],A[4],l);
        long long ans = 0;
        long long  f=1;
        for (int i = 1; i<=len ; i++){
            if ((len-i) % 2 ) f = -1LL;
            else f = 1LL;
            ans = (ans + f * F[i] * A[1][i] % MOD + MOD )%MOD;
        }
        cout << (ans % MOD ) << endl;
    }
}

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