[SDOI2017]序列计数

嘟嘟嘟


这题有那么难？？？？
提供一个吊打std的作法


直接令\(dp[i][j][0/1]\)表示前\(i\)个数的和模\(p\)为\(j\)，且这\(i\)个数中没有/有质数的方案数。
先想一下暴力，枚举第\(i\)个数是哪个，而后根据这个数是不是质数转移便可。复杂度\(O(nmp)\)。


优化：
发现\(n \leqslant 10 ^ 9\)，就能想到多项式快速幂，转移的时候分四种状况讨论。由于\(p\)很小，暴力乘就能过，复杂度\(O(m + p^ 2 logn)\)（\(O(m)\)是筛质数复杂度）。


这 题 就 没 了。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<vector>
#include<ctime>
#include<assert.h>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; (y = e[i].to) && ~i; i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxs = 105;
const int maxm = 2e7 + 5;
const int maxp = 105;
const ll mod = 20170408; 
In ll read()
{
    ll ans = 0;
    char ch = getchar(), las = ' ';
    while(!isdigit(ch)) las = ch, ch = getchar();
    while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
    if(las == '-') ans = -ans;
    return ans;
}
In void write(ll x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
    freopen("ha.in", "r", stdin);
    freopen("ha.out", "w", stdout);
#endif
}

int n, m, p;

In ll inc(ll a, ll b) {return a + b < mod ? a + b : a + b - mod;}

int prm[maxm], v[maxm];
In void init()
{
    for(int i = 2; i < maxm; ++i)
    {
        if(!v[i]) v[i] = i, prm[++prm[0]] = i;
        for(int j = 1; j <= prm[0] && i * prm[j] < maxm; ++j)
        {
            v[i * prm[j]] = prm[j];
            if(i % prm[j] == 0) break;
        }
    }
}

ll dp[maxs][maxp][2];
In void work0()
{
    dp[0][0][0] = 1;
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 0; j < p; ++j)
            for(int k = 1; k <= m; ++k) 
            {
                int t = (j + k) % p;
                if(v[k] ^ k) dp[i][t][0] = inc(dp[i][t][0], dp[i - 1][j][0]);
                dp[i][t][1] = inc(dp[i][t][1], dp[i - 1][j][1]);
                if(v[k] == k) dp[i][t][1] = inc(dp[i][t][1], dp[i - 1][j][0]);
            }
    }
    write(dp[n][0][1]), enter;
}

ll f[maxp][2], g[maxp][2], c[maxp][2];
In void mul(ll a[][2], ll b[][2], bool flg)
{
    Mem(c, 0);
    for(int i = 0; i < p; ++i)
        for(int j = 0; j < p; ++j)
        {
            int t = i + j < p ? i + j : i + j - p;
            c[t][0] = inc(c[t][0], a[i][0] * b[j][0] % mod);
            c[t][1] = inc(c[t][1], a[i][1] * b[j][0] % mod);
            c[t][1] = inc(c[t][1], a[i][0] * b[j][1] % mod);
            c[t][1] = inc(c[t][1], a[i][1] * b[j][1] % mod);
        }
    memcpy(a, c, sizeof(c));
}
In ll Quickpow(int n)
{
    f[0][0] = 1;
    for(int i = 1; i <= m; ++i) ++g[i % p][v[i] == i];
    for(; n; n >>= 1, mul(g, g, 0))
        if(n & 1) mul(f, g, 1);
    return f[0][1];
}

int main()
{
//  MYFILE();
    init();
    n = read(), m = read(), p = read();
    if(n <= 100) {work0(); return 0;}
    write(Quickpow(n)), enter;
    return 0;
}