Bash Plays with Functions CodeForces

大意: 定义函数$f_r(n)$, $f_0(n)$为pq=n且gcd(p,q)=1的有序对(p,q)个数.

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

$q$组询问, 求$f_r(n)$的值模1e9+7.

显然可以得到$f_0(n)=2^{\omega(n)}$, 是积性函数.

所以$f_r=f_{r-1}*1$也为积性函数, 然后积性函数$dp$即可.

问题就转化为对每个素数$p$, 求$dp[p][r][k]=f_r(p^k)$.

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

而$dp[p][0][k]=1$, 所以每个素数贡献相同, 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);

	}

}

巴特西

Bash Plays with Functions CodeForces - 757E (积性函数dp)

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