2017 Multi-University Training Contest - Team 3 RXD and functions（NTT）

题解：

我是参考的

http://blog.csdn.net/qq_32570675/article/details/76571666

这一篇

orz 原来可以这么变换，涨姿势

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + ;
const int g = ;
const int MOD = ;
LL F[maxn*], G[maxn*], mul[maxn], imul[maxn], ans[maxn], qp[maxn];
int a[maxn];

LL mypow(LL a, LL b){
    LL ans = ; for(; b; b >>= ) { if(b&) (ans *= a) %= MOD; (a *= a) %= MOD; } return ans;
}

void Modify(LL *a, int L){
    for(int i = , j = L/; i < L-; i++){
        if(i < j) swap(a[i], a[j]);
        int k = L/;
        while(j >= k){
            j -= k;
            k >>= ;
        }
        if(j < k) j += k;
    }
}

void NTT(LL *y, int L, int ty){
    Modify(y, L);
    int id = ;
    for(int h = ; h <= L; h <<= ){
        id++;
        for(int j = ; j < L; j += h){
            LL w = ;
            for(int k = j; k < j + h/; k++){
                LL u = y[k];
                LL t = y[k+h/]*w%MOD;
                y[k] = (u+t)%MOD;
                y[k+h/] = (u-t+MOD)%MOD;
                w = w*qp[id]%MOD;
            }
        }
    }
    if(ty == -){
        for(int i = ; i < L/; i++) swap(y[i], y[L-i]);
        LL inv = mypow(L, MOD-);
        for(int i = ; i < L; i++)
            y[i] = y[i]*inv%MOD;
    }
}

void init(){
    int n = 1e5 + ;
    mul[] = imul[] = mul[] = ;
    for(int i = ; i <= n; i++) mul[i] = (mul[i-]*i)%MOD;
    for(int i = ; i <= n; i++) imul[i] = mypow(mul[i], MOD-);
    for(int i = ; i < ; i++){
        int t = <<i;
        qp[i] = mypow(g, (MOD-)/t);
    }
}

int main()
{
    init();
    int n, m, x;
    while(cin>>n){
        for(int i = ; i <= n; i++) scanf("%d", &a[i]);
        cin>>m;
        LL b = ;
        for(int i = ; i <= m; i++){
            scanf("%d", &x);
            b -= x;
            (b += MOD) %= MOD;
        }
        memset(F, , sizeof(F));
        memset(G, , sizeof(G));
        for(int i = ; i <= n; i++){
            F[i] = a[n-i]*mul[n-i]%MOD;
            G[i] = mypow(b, i)*imul[i]%MOD;
        }
        int L = , nn = (n+)<<;
        while(L < nn) L <<= ;
        NTT(F, L, ); NTT(G, L, );
        for(int i = ; i < L; i++){
            F[i] = (F[i]*G[i])%MOD;
        }
        NTT(F, L, -);
        for(int i = ; i <= n; i++){
            (F[i] += MOD) %= MOD;
            ans[i] = F[n-i]*imul[i]%MOD;
        }
        for(int i = ; i <= n; i++) printf("%d ", ans[i]);
        cout<<endl;
    }
    return ;
}