约数相关：约数个数

N=(p1^c1)*(p2c2)...(pk^ck)
N^2=(p1(c1**2)) * (p2^ (c22) )...(pk^ (ck2) )

约数个数 f[N]=(c1+1)(c2+1)...(cn+1)

拍打牛头https://www.acwing.com/problem/content/1293/

这里没有用到公式 只是将求约数转化成为求倍数

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

using namespace std;

const int N = 1000010;

int n;
int a[N], cnt[N], s[N];

int main()
{
    scanf("%d", &n);

    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &a[i]);
        cnt[a[i]] ++ ;//计数
    }

    for (int i = 1; i < N; i ++ )
        for (int j = i; j < N; j += i)//倍数都加一遍 包括自己
            s[j] += cnt[i];

    for (int i = 0; i < n; i ++ ) printf("%d\n", s[a[i]] - 1);

    return 0;
}

标签：约数,...,cnt,int,个数,++,相关,include
来源： https://www.cnblogs.com/liang302/p/16558235.html