洛谷P4245 【模板】MTT(任意模数NTT)

题目背景

模板题，无背景

题目描述

给定

系数对 $\cdot 2^k + 1p=a⋅2k+1 之形式。$

输入输出格式

输入格式：

输入共 $a_iai 表示 F(x)F(x) 的 i-1i−1 次项的系数。第三行为 m+1m+1 个整数，第 ii 个整数 b_ibi 表示 G(x)G(x) 的 i-1i−1 次项的系数。$

输出格式：

输出 $c_ici 表示 F(x) * G(x)F(x)∗G(x) 的 i-1i−1 次项的系数。$

输入输出样例

输入样例#1：复制

5 8 28
19 32 0 182 99 95
77 54 15 3 98 66 21 20 38

输出样例#1：复制

7 18 25 19 5 13 12 2 9 22 5 27 6 26

说明

$\leq n \leq 10^5, 0 \leq a_i, b_i \leq 10^9, 2 \leq p \leq 10^9 + 91≤n≤105,0≤ai,bi≤109,2≤p≤109+9$

MTT不会，

只好用三模数NTT搞

板子题

原理可以看这里

真TM恶心。。

#include<cstdio>
#include<algorithm>
#include<cstring>
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<21, stdin), p1 == p2) ? EOF : *p1++)
#define swap(x,y) x ^= y, y ^= x, x ^= y
#define LL long long 
const int MAXN = 3 * 1e6 + 10;
using namespace std;
char buf[1<<21], *p1 = buf, *p2 = buf;
inline int read() { 
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
const int P1 = 469762049, P2 = 998244353, P3 = 1004535809, g = 3; 
const LL PP = 1ll * P1 * P2;
int N, M, P, limit = 1, L;
int A[MAXN], B[MAXN], C[MAXN], D[MAXN], Ans[3][MAXN], r[MAXN];
LL fastmul(LL a, LL b, LL mod) {
    a %= mod, b %= mod;
    return ((a * b - (LL)((LL)((long double)a / mod * b + 1e-3) * mod)) % mod + mod) % mod;
}
int fastpow(int a, int p, int mod) {
    int base = 1;
    while(p) {
        if(p & 1) base = 1ll * a * base % mod;
        a = 1ll * a * a % mod; p >>= 1;
    }
    return base % mod;
} 
void NTT(int *A, const int n, const int type, const int mod) {
    for(int i = 0; i < n; i++)
        if(i < r[i]) swap(A[i], A[r[i]]);
    for(int mid = 1; mid < n; mid <<= 1) {
        int W = fastpow(type == 1 ? g : fastpow(g, mod - 2, mod) , (mod - 1) / (mid << 1), mod);
        for(int j = 0; j < n; j += (mid << 1)) {
            int w = 1;
            for(int k = 0; k <mid; k++, w = 1ll * w * W % mod) {
                int x = A[j + k], y = 1ll * w * A[j + k + mid] % mod;
                A[j + k] = (x + y) % mod,
                A[j + k + mid] = (x - y + mod) % mod;
            }
        }
    }
    if(type == -1) {
        int inv = fastpow(n, mod - 2, mod);
        for(int i = 0; i < n; i++) 
            A[i] = 1ll * A[i] * inv % mod;
    }
}

int main() {
    #ifdef WIN32
    freopen("a.in", "r", stdin);
    #endif
    N = read(), M = read(), P = read();
    for(int i = 0; i <= N; i++) A[i] = read();
    for(int i = 0; i <= M; i++) B[i] = read();
    
    while(limit <= N + M) limit <<= 1, L++;
    for(int i = 0; i <= limit; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
    
    copy(A, A + N + 1, C); copy(B, B + M + 1, D);
    NTT(C, limit, 1, P1); NTT(D, limit, 1, P1);
    for(int i = 0; i <= limit; i++) Ans[0][i] = 1ll * C[i] * D[i] % P1;
    
    memset(C, 0, sizeof(C)); memset(D, 0, sizeof(D));
    copy(A, A + N + 1, C); copy(B, B + M + 1, D);
    NTT(C, limit, 1, P2); NTT(D, limit, 1, P2);
    for(int i = 0; i <= limit; i++) Ans[1][i] = 1ll * C[i] * D[i] % P2;
    
    memset(C, 0, sizeof(C)); memset(D, 0, sizeof(D));
    copy(A, A + N + 1, C); copy(B, B + M + 1, D);
    NTT(C, limit, 1, P3); NTT(D, limit, 1, P3);
    for(int i = 0; i <= limit; i++) Ans[2][i] = 1ll * C[i] * D[i] % P3;
    
    NTT(Ans[0], limit, -1, P1);
    NTT(Ans[1], limit, -1, P2);
    NTT(Ans[2], limit, -1, P3);
    
    for(int i = 0; i <= N + M; i++) {
        LL A = (fastmul(1ll * Ans[0][i] * P2 % PP, fastpow(P2 % P1, P1 - 2, P1), PP) + 
                fastmul(1ll * Ans[1][i] * P1 % PP, fastpow(P1 % P2, P2 - 2, P2), PP) ) % PP;
        LL K = ((Ans[2][i] - A) % P3 + P3) % P3 * fastpow(PP % P3, P3 - 2, P3) % P3;
        printf("%d ",(A % P + ((K % P) * (PP % P)) % P ) % P);         
    }
    return 0;
}

洛谷P4245 【模板】MTT(任意模数NTT)

题目背景

题目描述

输入输出格式

输入输出样例

说明

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