loj#2002. 「SDOI2017」序列计数(dp 矩阵乘法)

题意

sol

质数的限制并没有什么卵用，直接容斥一下：答案 = 忽略质数总的方案 - 没有质数的方案

那么直接dp，设\(f[i][j]\)表示到第i个位置，当前和为j的方案数

\(f[i + 1][(j + k) \% p] += f[i][j]\)

矩乘优化一下。

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int maxn = 2e7 + 10, mod = 20170408, ss = 1e5 + 10;
ll gg = 1e17;
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;
}
template<typename a, typename b> inline int add(a x, b y) {
    if(x + y < 0) return x + y + mod;
    else return x + y >= mod ? x + y - mod : x + y;
}
template<typename a, typename b> inline void add2(a &x, b y) {
    if(x + y < 0) x = x + y + mod;
    else x = (x + y >= mod ? x + y - mod : x + y);
}
template<typename a, typename b> inline int mul(a x, b y) {
    return 1ll * x * y % mod;
}
int n, m, p, lim;//1 - m, ºíêçpµä±¶êý 
int f[ss], vis[maxn], mu[maxn], prime[maxn], tot, cnt, num[ss], tim[ss], val[ss];
struct ma {
    int m[201][201];
    ma() {
        memset(m, 0, sizeof(m));
    }
    void init() {
        for(int i = 0; i <= lim; i++) m[i][i] = 1;
    }
    void clear() {
        memset(m, 0, sizeof(m));
    }
    void print() {
        for(int i = 0; i <= lim; i++, puts(""))
            for(int j = 0; j <= lim; j++)
                printf("%d ", m[i][j]);
    }
    ma operator * (const ma &rhs) const {
        ma ans = {};
        for(int i = 0; i <= lim; i++)
            for(int j = 0; j <= lim; j++) {
                __int128 tmp = 0;
                for(int k = 0; k <= lim; k++) {
                    tmp += 1ll * m[i][k] * rhs.m[k][j];     
                }
                ans.m[i][j] = tmp % mod;
            }
        return ans;
    }
}g;
ma matrixpow(ma a, int p) {
    ma base; base.init();
    while(p) {
        if(p & 1) base = base * a;
        a = a * a; p >>= 1;
    }
    return base;
}
void sieve(int n) {
    vis[1] = 1; mu[1] = 1; 
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) prime[++tot] = i, mu[i] = -1;
        for(int j = 1; j <= tot && i * prime[j] <= n; j++) {
            vis[i * prime[j]] = 1;
            if(i % prime[j]) mu[i * prime[j]] = -mu[i];
            else {mu[i * prime[j]] = 0; break;}
        }
    }
    for(int i = 1; i <= n; i++) 
        if(vis[i]) num[i % p]++;
}

int solve1() {//ºöêóöêêýµäïþöæ
    for(int i = 1; i <= m; i++) f[i % p]++;
    for(int j = 0; j < p; j++) {
        memset(tim, 0, sizeof(tim));
        memset(val, 0, sizeof(val));
        int step = m;
        for(int k = 1; k <= m; k++) {
            int nxt = (j + k) % p;
            if(tim[nxt]) {step = k - 1; break;}
            tim[nxt] = 1; val[nxt]++;
        }
        if(step) for(int k = 0; k <= lim; k++) g.m[k][j] = m / step * val[k];
        for(int k = m / step * step + 1; k <= m; k++) g.m[(j + k) % p][j]++;
    }
    ma ans = matrixpow(g, n - 1);
    int out = 0;
    for(int i = 0; i <= lim; i++) add2(out, mul(ans.m[0][i], f[i]));
    return out;
}
int solve2() {//îþöêêý 
    memset(f, 0, sizeof(f));
    g.clear();
    for(int i = 1; i <= m; i++) f[i % p] += (vis[i]);
    for(int j = 0; j < p; j++)
        for(int k = 0; k < p; k++)
            g.m[(j + k) % p][j] += num[k];
            
    ma ans = matrixpow(g, n - 1);
    int out = 0;
    for(int i = 0; i <= lim; i++) 
        add2(out, mul(ans.m[0][i], f[i]));
    return out;
}
int main() {
    n = read(); m = read(); lim = p = read();
    sieve(m);
    cout << (solve1() - solve2() + mod) % mod;
    return 0;
}