大意: 定義函數$f_r(n)$, $f_0(n)$為pq=n且gcd(p,q)=1的有序對(p,q)個數.

$r \ge 1$時, $f_r(n)=\sum\limits_{uv=n}\frac{f_{r-1}(u)+f_{r-1}(v)}{2}$.

$q$組詢問, 求$f_r(n)$的值模1e9+7.

顯然可以得到$f_0(n)=2^{\omega(n)}$, 是積性函數.

所以$f_r=f_{r-1}*1$也為積性函數, 然後積性函數$dp$即可.

問題就轉化為對每個素數$p$, 求$dp[p][r][k]=f_r(p^k)$.

$dp[p][r][k]=\sum\limits_{x=0}^k dp[p][r-1][x]$.

而$dp[p][0][k]=1$, 所以每個素數貢獻相同, 只需要$dp$一次即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head



const int N = 1e6+10;
int dp[N][22], sum[N][22], mi[N];

int main() {
	dp[0][0]=sum[0][0]=1;
	REP(i,1,21) sum[0][i]=sum[0][i-1]+(dp[0][i]=2);
	REP(i,1,N-1) {
		dp[i][0]=sum[i][0]=1;
		REP(j,1,21) sum[i][j]=(sum[i][j-1]+(dp[i][j]=sum[i-1][j]))%P;
	}
	REP(i,1,N-1) mi[i] = i;
	REP(i,2,N-1) if (mi[i]==i) {
		for (int j=i; j<N; j+=i) mi[j]=min(mi[j],i);
	}
	int q;
	scanf("%d", &q);
	while (q--) {
		int r, n;
		scanf("%d%d", &r, &n);
		int ans = 1;
		while (n!=1) {
			int t = mi[n], k = 0;
			while (n%t==0) n/=t, ++k;
			ans = (ll)ans*dp[r][k]%P;
		}
		printf("%d\n", ans);
	}
}

Bash Plays with Functions CodeForces - 757E (積性函數dp)