正着求sum会WA两个点

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>

using namespace std;

const int N = 1000007;
typedef long long ll;
ll mod = 1e9 + 7;

ll n, m, cnt;
int primes[N];
bool vis[N];

void get_primes(ll n)
{
    vis[0] = vis[1] = 1;
    for(int i = 2; i <= n; ++ i){
        if(vis[i] == 0){
            primes[++ cnt] = i;
        }
        for(int j = 1; j <= cnt && i * primes[j] <= n; ++ j){
            vis[i * primes[j]] = true;
            if(i % primes[j] == 0)break;
        }
    }
}

ll qpow(ll a, ll b)
{
    ll res = 1;
    while(b){
        if(b & 1)res = (res * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return res;
}

int main(){
    scanf("%lld", &n);
    get_primes(n);
    ll ans = 1;
    for(int i = 1; i <= cnt; ++ i){
        int p = primes[i];
        ll sum = 0;
        for(int j = 1; qpow(p, j) <= n; ++ j){
            sum = sum + n / qpow(p, j);
        }
        ans = ans * (2 * sum + 1) % mod;
    }
    printf("%lld\n", ans % mod);
    return 0;
}

倒着求sum就AC了

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>

using namespace std;

const int N = 1000007;
typedef long long ll;
ll mod = 1e9 + 7;

ll n, m, cnt;
int primes[N];
bool vis[N];

void get_primes(ll n)
{
    vis[0] = vis[1] = 1;
    for(int i = 2; i <= n; ++ i){
        if(vis[i] == 0){
            primes[++ cnt] = i;
        }
        for(int j = 1; j <= cnt && i * primes[j] <= n; ++ j){
            vis[i * primes[j]] = true;
            if(i % primes[j] == 0)break;
        }
    }
}

int main(){
    scanf("%lld", &n);
    get_primes(n);
    ll res = 1;
    for(int i = 1; i <= cnt; ++ i){
        int p = primes[i];
        int sum = 0;
        for(int j = n; j; j /= p)
            sum += j / p;
        res = (res * (2 * sum + 1)) % mod;
    }
    cout << res % mod << endl;
    return 0;
}

for(int j = 1; qpow(p, j) <= n; ++ j)
	sum = sum + n / qpow(p, j);

for(int j = n; j; j /= p)
	sum += j / p;

这两种有什么区别嘛