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codeforces757E. Bash Plays with Functions(狄…

2018-07-18 01:15:41来源：博客园阅读 ()

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 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
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;
}
int prime[MAXN], tot, vis[MAXN];
LL f[MAXN][22];
void GetPrime(int N) {
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i; 
        for(int j = 1; j <= N && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
        }
    }
}
void Pre(int N, int M) {
    f[0][0] = 1;//f[i][k] f_r(p^k)
    for(int i = 1; i <= M; i++) f[0][i] = 2;
    for(int r = 1; r <= N; r++) {
        LL sum = 0;
        for(int k = 0; k <= M; k++) {
            sum += f[r - 1][k];
            (f[r][k] += sum ) %= mod;
        }
    }
}
main() {
    GetPrime(1e6 + 5);
    Pre(1e6 + 5, 21);
    int Q = read();
    while(Q--) {
        int r = read(), n = read();
        LL ans = 1;
        for(int i = 1; i <= tot && prime[i] <= sqrt(n); i++) {
            if(n % prime[i]) continue;
            int num = 0;
            while(!(n % prime[i])) num++, n /= prime[i];
            ans = 1ll * ans * (f[r][num]) % mod;
        }
        if(n > 1) ans = (1ll * ans * f[r][1]) % mod;
        printf("%I64d\n", ans);
    }
}
/*

*/