CodeForces - 58B Coins（数论）

最新推荐文章于 2019-09-14 10:50:07 发布

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#CodeForces - 58B #Coins

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本文探讨了一种新的货币系统设计问题，在该系统中，最昂贵的硬币面额为n单位伯兰币，且每种硬币的面额必须能被更便宜的硬币整除。文章提供了一个算法解决方案，通过分解质因数来确定最优硬币面额配置。

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Coins

In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.

Input

The first and only line contains an integer n (1 ≤ n ≤ 106) which represents the denomination of the most expensive coin.

Output

Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.

    Examples
  
      input
    
10

      output
    
10 5 1

      input
    
4

      output
    
4 2 1

      input
    
3

      output
    
3 1

ps：其实就是分解质因子，并依次除以质因子就好了（线性筛法想了解的请戳这里）

代码：

#include<stdio.h>

#define maxn 1000000+10
#define max(a,b) (a>b?a:b)
int p[maxn/10],c[maxn];

void is_prim()//线性筛法
{
    int tot=0,i,j;
    c[0]=1,c[1]=1;
    for(i=2; i<=maxn; i++)
    {
        if(!c[i])
            p[tot++]=i;
        for(j=0; j<tot&&i*p[j]<maxn; j++)
        {
            c[i*p[j]]=1;
            if(!(i%p[j]))
                break;
        }
    }
}

int main()
{
    is_prim();
    int n;
    scanf("%d",&n);
    printf("%d ",n);
    int ps=0;
    while(n!=1)
    {
        if((n%p[ps])==0)
        {
            n/=p[ps];
            printf("%d ",n);
        }
        else
            ps++;
    }
    return 0;
}