joj 2344: Dissolution of Integer 将N分解成乘积形式共有多少种解法

kongming_acm

于 2011-10-04 12:05:05 发布

阅读量711

点赞数

CC 4.0 BY-SA版权

分类专栏： joj acm_数学问题 joj acm_动态规划文章标签： integer forms output input each

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/6843690

joj acm_数学问题同时被 2 个专栏收录

63 篇文章

订阅专栏

joj acm_动态规划

29 篇文章

订阅专栏

本文介绍了一种用于计算正整数不同分解形式数量的高效算法，并提供了完整的C++实现代码。该算法通过预处理得到每个数的分解方式数量，进而快速响应查询请求。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

As you known,a positive integer can be written to a form of products of several positive integers(N = x1 * x2 * x3 *.....* xm (xi!=1 (1<=i<=m))).Now, given a positive integer, please tell me how many forms of products.

Input

Input consists of several test cases.Given a positive integer N (2<=N<=200000) in every case.

Output

For each case, you should output how many forms of the products.

Sample Input

12
100
5968

Sample Output

4
9
12

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=200001;
int a[maxn],prime[maxn],pl;
int main()
{
    a[1]=1;
    for(int i=2;i<maxn;i++)
    {
        for(int j=1;i*j<maxn;j++)
        {
            a[i*j]+=a[j];
        }
    }
    int n;
    while(scanf("%d",&n)==1)
    {
        printf("%d\n",a[n]);
    }
    return 0;
}