Codeforces 757E 积性函数

1.首先 百度学习积性函数定义 或者 wiki学multiplicative function
2.$f_{r+1}(n) = \sum_{d|n}f_{r}(d)$ 这个还是很好发现的
3.假设$n = p_{1}^{e1}p_{2}^{e2}$
4.因为$f_{0}$ 是积性函数，所以$f_{0}(n)= f_{0}(p_{1}^{e1}) * f_{0}(p_{2}^{e2})$
5.

$$f_{1}(n) = \sum_{d|n}f_{0}(d) = \sum_{i=0}^{e1}\sum_{j=0}^{e2}f_{0}(p_{1}^{i}p_{2}^{j}) = \sum_{i=0}^{e1}f_{0}(p_{1}^{i}) * \sum_{j=0}^{e2}f_{0}(p_{2}^{j}) = f_{1}(p_{1}^{e1})*f_{1}(p_{2}^{e2})$$
所以 $f_{1}$是积性函数
同理 $f_{r+1}$也是积性函数
6.故要求的就是 $f_{r+1}(p^{e}) = f_{r+1}(p^{e-1})+f_{r}(p^{e})$而且这个值和p是多少是没有关系的，开个dp推一下就行了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <cctype>
#include <cmath>
#include <vector>
#include <sstream>
#include <bitset>
#include <deque>
#include <iomanip>
using namespace std;
#define pr(x) cout << #x << " = " << x << endl;
#define bug cout << "bugbug" << endl;
#define ppr(x, y) printf("(%d, %d)\n", x, y);
#define MST(a,b) memset(a,b,sizeof(a))
#define CLR(a) MST(a,0)
#define SQR(a) ((a)*(a))
#define PCUT puts("\n---------------")

typedef long long ll;
typedef double DBL;
typedef pair<int, int> P;
typedef unsigned int uint;
const int MOD = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 1e6 + 4;
const int maxm = 2e1 + 4;
const double pi = acos(-1.0);
ll dp[maxn][maxm];
vector<int> fac[maxn];

void divide(int num){
    int t = num;
    for (int i = 2; i * i <= num; ++i)
        if (num % i == 0){
            int tmp = 0;
            while(num % i == 0) tmp++, num /= i;
            fac[t].push_back(tmp); 
        }
    if (num != 1){
        fac[t].push_back(1);
    }
    return;
} 

void init(){
    for (int i = 0; i < maxm; ++i) dp[0][i] = 1 << i;
    for (int i = 0; i < maxm; ++i) dp[1][i] = 1 + 2 * i; 
    for (int i = 2; i < maxn; ++i){
        dp[i][0] = 1;
        for (int j = 1; j < maxm; ++j) dp[i][j] = (dp[i][j-1] + dp[i-1][j]) % MOD;
    }
    return;
}
ll fun(int r, int num){
    if (r == 0) return dp[0][fac[num].size()];
    ll sum = 1;
    for (int i = 0; i < fac[num].size(); ++i){
        sum = sum * dp[r][fac[num][i]] % MOD; 
//      cout << fac[num][i] << ' '; 
    } 

    return sum;
}
int main(){
//必须编译过才能交
    int ik, i, j, k, kase;
    init();
    for (int i = 1; i < maxn; ++i) divide(i);
    int q;
    scanf("%d", &q);
    while(q--){
        int r, num;
        scanf("%d%d", &r, &num);
        printf("%I64d\n", fun(r, num));
    }
    return 0;
}