P3702 [SDOI2017]序列计数 (三模数NTT)

题意：

Alice想要得到一个长度为$n$的序列，序列中的数都是不超过$m$的正整数，而且这$n$个数的和是$n$的倍数。

Alice还希望，这$n$个数中，至少有一个数是质数。

Alice想知道，有多少个序列满足她的要求。

思路：
利用容斥原理，这个题目其实就是

　在集合S(1~m)中选n个数的和模p为0的方案数-在集合S(1~m的合数)中选n个数的和模p为0的方案数(类似模型：洛谷P3321)

Code:

#include <map>
#include <set>
#include <array>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)
#define m1 998244353
#define m2 469762049
#define m3 1004535809

const int N = 3e2 + 10;
const int M = 2e7 + 20;
const int mod = 20170408;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int Mod[] = {998244353, 469762049, 1004535809};

int n, m, p;
int rev[N];
ll vis[N], h[N];
int primes[M], cnt = 0;
bool st[M];

void get(int n)
{
    st[1] = true;
    for (int i = 2; i <= n; i++)
    {
        if (!st[i])
            primes[cnt++] = i;
        for (int j = 0; primes[j] <= n / i; j++)
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0)
            {
                break;
            }
        }
    }
}

ll qmi(ll a, ll b, ll p)
{
    ll res = 1;
    while (b)
    {
        if (b & 1)
            res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}

void change(ll y[], int len)
{
    for (int i = 0; i < len; i++)
    {
        rev[i] = rev[i >> 1] >> 1;
        if (i & 1)
            rev[i] |= (len >> 1);
    }

    for (int i = 0; i < len; i++)
    {
        if (i < rev[i])
            swap(y[i], y[rev[i]]);
    }
}

void ntt(ll y[], int len, int on, ll MOD)
{
    change(y, len);
    for (int h = 2; h <= len; h <<= 1)
    {
        ll wn = qmi(3, (MOD - 1) / h, MOD);
        if (on == -1)
            wn = qmi(wn, MOD - 2, MOD);

        for (int j = 0; j < len; j += h)
        {
            ll w = 1;
            for (int k = j; k < j + h / 2; k++)
            {
                ll u = y[k];
                ll t = w * y[k + h / 2] % MOD;
                y[k] = (u + t) % MOD;
                y[k + h / 2] = (u - t + MOD) % MOD;
                w = w * wn % MOD;
            }
        }
    }

    if (on == -1)
    {
        ll inv = qmi(len, MOD - 2, MOD);
        for (int i = 0; i < len; i++)
        {
            y[i] = y[i] * inv % MOD;
        }
    }
}

ll mult(ll a, ll b, ll p)
{
    ll res = 0;
    while (b)
    {
        if (b & 1)
            res = (res + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return res;
}

ll A[N], B[N], C[N], D[N];
void mul(ll a[], ll b[], ll res[], ll len)
{
    memcpy(A, a, sizeof(A));
    memcpy(B, a, sizeof(B));
    memcpy(C, a, sizeof(C));
    memcpy(D, b, sizeof(D));

    ntt(A, len, 1, Mod[0]);
    ntt(D, len, 1, Mod[0]);
    for (int i = 0; i < len; i++)
    {
        A[i] = A[i] * D[i] % Mod[0];
    }
    ntt(A, len, -1, Mod[0]);

    memcpy(D, b, sizeof(D));
    ntt(B, len, 1, Mod[1]);
    ntt(D, len, 1, Mod[1]);
    for (int i = 0; i < len; i++)
    {
        B[i] = B[i] * D[i] % Mod[1];
    }
    ntt(B, len, -1, Mod[1]);

    memcpy(D, b, sizeof(D));
    ntt(C, len, 1, Mod[2]);
    ntt(D, len, 1, Mod[2]);
    for (int i = 0; i < len; i++)
    {
        C[i] = C[i] * D[i] % Mod[2];
    }
    ntt(C, len, -1, Mod[2]);

    ll M12 = 1ll * m1 * m2;
    ll inv2 = qmi(m2, m1 - 2, m1);
    ll inv1 = qmi(m1, m2 - 2, m2);
    ll mul2 = 1ll * m2 * inv2 % M12;
    ll mul1 = 1ll * m1 * inv1 % M12;
    ll inv = qmi(M12 % m3, m3 - 2, m3);
    ll m12 = M12 % mod;
    ll c1, c2, c3, c4, q;

    for (int i = 0; i <= (p << 1); i++)
    {
        c1 = A[i], c2 = B[i], c3 = C[i];
        c4 = (mult(c1, mul2, M12) + mult(c2, mul1, M12)) % M12;
        q = ((c3 - c4) % m3 + m3) % m3 * inv % m3;
        res[i] = (q * m12 % mod + c4) % mod;
    }
    for (int i = p; i < len; i++)
    {
        res[i % p] = (res[i % p] + res[i]) % mod;
        res[i] = 0;
    }
}

ll res[N];

void qmi_ntt(ll y[], int len, int n)
{
    memset(res, 0, sizeof(res));
    res[0] = 1;
    while (n)
    {
        if (n & 1)
        {
            mul(res, y, res, len);
        }
        mul(y, y, y, len);
        n >>= 1;
    }
}

ll mid[N], ans[3], ans1, ans2;

void solve()
{
    cin >> n >> m >> p;

    get(m);
    for (int i = 1; i <= m; i++)
    {
        vis[i % p]++;
        if (st[i])
            h[i % p]++;
    }

    int len = 1;
    while (len <= p + p - 1)
        len <<= 1;

    qmi_ntt(vis, len, n);

    ans1 = res[0];

    qmi_ntt(h, len, n);

    ans1 = (ans1 - res[0] + mod) % mod;
    cout << ans1 << "\n";
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/桌面/11.21/2/in9.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    // sd(T);
    // cin >> T;
    while (T--)
    {
        solve();
    }

    return 0;
}

标签：int,P3702,ll,NTT,len,ntt,SDOI2017,include,Mod
来源： https://www.cnblogs.com/jungu/p/14416333.html