AcWing 871. 约数之和（唯一分解定理+组合数）

Posted 2022-02-20 MangataTS

题目连接

https://www.acwing.com/problem/content/description/873/

思路

如果n可以写成$f(n)=a_1^{{\alpha }_1}\times a_2^{{\alpha }_2}\dots \dots \times a_k^{{\alpha }_k}$（$a_1,a_2,\dots \dots ,a_k$均为质数）
那么n的约数的和为：
$sum=(a_1^0+a_1^1+a_1^2+\dots \dots a_1^{{\alpha }_1})\times (a_2^0+a_2^1+a_2^2+\dots \dots a_2^{{\alpha }_2})\times \dots \dots \times (a_k^0+a_k^1+a_k^2+\dots \dots a_k^{{\alpha }_k})$
所以我们直接将每一个因子唯一分解后将数量统计出来，最后然后遍历将每个质因子的贡献通过求和公式算出并累乘即可，详情请看吗代码

代码

#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define ll long long
#define int long long
map<ll,ll> vis;
ll n,a;
void slove()
    for(ll i = 2;i * i <= a; ++i) 
        while(a % i == 0) vis[i]++,a /= i;
    
    if(a-1) vis[a]++;


ll ksm(ll a,ll b)
    ll ans = 1;
    while(b)
        if(b & 1) ans = (ans * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    
    return ans;


ll inv(ll x) return ksm(x,mod-2);

signed main()

    scanf("%lld",&n);
    for(ll i = 1;i <= n; ++i) 
        scanf("%lld",&a);
        slove();
    
    long long ans = 1LL;
    for(auto it : vis)  
        ll q = it.first;
        ll k = it.second;
        ll res = 0;
        if(q == 1) 
            res = n + 1;
        else 
            res = ((ksm(q,k + 1) - 1LL + mod) * inv(q-1)) % mod;
        
        ans = (ans * res) % mod;
    
    printf("%lld\\n",ans);

    return 0;