2017 Multi-University Training Contest

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 + 100;
const int g = 3;
const int MOD = 998244353;
LL F[maxn*8], G[maxn*8], mul[maxn], imul[maxn], ans[maxn], qp[maxn];
int a[maxn];

LL mypow(LL a, LL b){
    LL ans = 1; for(; b; b >>= 1) { if(b&1) (ans *= a) %= MOD; (a *= a) %= MOD; } return ans;
}

void Modify(LL *a, int L){
    for(int i = 1, j = L/2; i < L-1; i++){
        if(i < j) swap(a[i], a[j]);
        int k = L/2;
        while(j >= k){
            j -= k;
            k >>= 1;
        }
        if(j < k) j += k;
    }
}

void NTT(LL *y, int L, int ty){
    Modify(y, L);
    int id = 0;
    for(int h = 2; h <= L; h <<= 1){
        id++;
        for(int j = 0; j < L; j += h){
            LL w = 1;
            for(int k = j; k < j + h/2; k++){
                LL u = y[k];
                LL t = y[k+h/2]*w%MOD;
                y[k] = (u+t)%MOD;
                y[k+h/2] = (u-t+MOD)%MOD;
                w = w*qp[id]%MOD;
            }
        }
    }
    if(ty == -1){
        for(int i = 1; i < L/2; i++) swap(y[i], y[L-i]);
        LL inv = mypow(L, MOD-2);
        for(int i = 0; i < L; i++)
            y[i] = y[i]*inv%MOD;
    }
}



void init(){
    int n = 1e5 + 10;
    mul[0] = imul[0] = mul[1] = 1;
    for(int i = 2; i <= n; i++) mul[i] = (mul[i-1]*i)%MOD;
    for(int i = 1; i <= n; i++) imul[i] = mypow(mul[i], MOD-2);
    for(int i = 0; i < 21; i++){
        int t = 1<<i;
        qp[i] = mypow(g, (MOD-1)/t);
    }
}

int main()
{
    init();
    int n, m, x;
    while(cin>>n){
        for(int i = 0; i <= n; i++) scanf("%d", &a[i]);
        cin>>m;
        LL b = 0;
        for(int i = 1; i <= m; i++){
            scanf("%d", &x);
            b -= x;
            (b += MOD) %= MOD;
        }
        memset(F, 0, sizeof(F));
        memset(G, 0, sizeof(G));
        for(int i = 0; i <= n; i++){
            F[i] = a[n-i]*mul[n-i]%MOD;
            G[i] = mypow(b, i)*imul[i]%MOD;
        }
        int L = 1, nn = (n+1)<<1;
        while(L < nn) L <<= 1;
        NTT(F, L, 0); NTT(G, L, 0);
        for(int i = 0; i < L; i++){
            F[i] = (F[i]*G[i])%MOD;
        }
        NTT(F, L, -1);
        for(int i = 0; i <= n; i++){
            (F[i] += MOD) %= MOD;
            ans[i] = F[n-i]*imul[i]%MOD;
        }
        for(int i = 0; i <= n; i++) printf("%d ", ans[i]);
        cout<<endl;
    }
    return 0;
}