codeforces757E. Bash Plays with Functions(狄利克雷卷积 积性函数)

http://codeforces.com/contest/757/problem/E

题意

Sol

非常骚的一道题

首先把给的式子化一下，设$u = d$，那么$v = n / d$

$$f_r(n) = \sum_{d \mid n} \frac{f_{r - 1}(d) + f_{r - 1}(\frac{n}{d})}{2}$$

$$= \sum_{d\mid n} f_{r - 1}(d)$$

很显然，这是$f_r(n)$与$1$的狄利克雷卷积

根据归纳法可以证明$f_r(n)$为积性函数

我们可以对每个质因子分别考虑他们的贡献

考虑$f_0(p^k) = [k =0]+1$，与$p$是无关的，因此我们只要枚举$r$和$k$就好

$f_r(p^k) = \sum_{i = 0}^k f_{r - 1}(p^i)$

前缀和优化dp

#include<cstdio>
#include<cmath>
#define LL long long
using namespace std;
const int MAXN  = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int prime[MAXN], tot, vis[MAXN];
LL f[MAXN][];
void GetPrime(int N) {
    for(int i = ; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i;
        for(int j = ; j <= N && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = ;
            if(i % prime[j] == ) break;
        }
    }
}
void Pre(int N, int M) {
    f[][] = ;//f[i][k] f_r(p^k)
    for(int i = ; i <= M; i++) f[][i] = ;
    for(int r = ; r <= N; r++) {
        LL sum = ;
        for(int k = ; k <= M; k++) {
            sum += f[r - ][k];
            (f[r][k] += sum ) %= mod;
        }
    }
}
main() {
    GetPrime(1e6 + );
    Pre(1e6 + , );
    int Q = read();
    while(Q--) {
        int r = read(), n = read();
        LL ans = ;
        for(int i = ; i <= tot && prime[i] <= sqrt(n); i++) {
            if(n % prime[i]) continue;
            int num = ;
            while(!(n % prime[i])) num++, n /= prime[i];
            ans = 1ll * ans * (f[r][num]) % mod;
        }
        if(n > ) ans = (1ll * ans * f[r][]) % mod;
        printf("%I64d\n", ans);
    }
}
/*

*/