UVa 10061 - How many zero's and how many digits ?

最新推荐文章于 2019-02-21 14:39:31 发布

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最新推荐文章于 2019-02-21 14:39:31 发布

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分类专栏： UVa 文章标签： system output each integer numbers input

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本文介绍了如何计算给定基数下的一个整数的阶乘的尾部零数量以及该阶乘的总位数，适用于不同进制系统。

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Problem G

How many zeros and how many digits?

Input: standard input

Output: standard output

Given a decimal integer number you will have to find out how many trailing zeros will be there in its factorial in a given number system and also you will have to find how many digits will its factorial have in a given number system? You can assume that for a b based number system there are b different symbols to denote values ranging from 0 ... b-1.

Input

There will be several lines of input. Each line makes a block. Each line will contain a decimal number N (a 20bit unsigned number) and a decimal number B (1<B<=800), which is the base of the number system you have to consider. As for example 5! = 120 (in decimal) but it is 78 in hexadecimal number system. So in Hexadecimal 5! has no trailing zeros

Output

For each line of input output in a single line how many trailing zeros will the factorial of that number have in the given number system and also how many digits will the factorial of that number have in that given number system. Separate these two numbers with a single space. You can be sure that the number of trailing zeros or the number of digits will not be greater than 2^31-1

Sample Input:

2 10
5 16
5 10

Sample Output:

0 1
0 2
1 3
________________________________________________________________________________________
Shahriar Manzoor
16-12-2000

建议大家看的博客地址为：http://www.cppblog.com/csu-yx/archive/2012/05/02/173512.aspx ，看了很多博客，感觉还是这个最好

#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
int main()
{
    int get_digit(int n,int m);
    int get_zero(int n,int m);
    unsigned int i,j,n,m,s,t;
    int k1,k2;
    while(cin>>n>>m)
    {
        k1=get_digit(n,m);
        k2=get_zero(n,m);
        cout<<k2<<" "<<k1<<endl;
    }
    return 0;
}
int get_digit(int n,int m)
{
    int s,i;
    double s1=0;
    for(i=1;i<=n;i++)
    {
        s1+=log10((double)i);
    }
    s=(int)(s1/log10((double)m)+0.01)+1;
    return s;
}
int get_zero(int n,int m)
{
    int a[1000],cal;
    int i,j;
    memset(a,0,sizeof(a));
    for(i=2;i<=n;i++)
    {
        cal=i;
        for(j=2;j<=cal&&j<=m;j++)
        {
            while(cal%j==0)
            {
                a[j]++;
                cal/=j;
            }
        }
    }
    int ans=0;
    while(1)
    {
        cal=m;
        for(i=2;i<=cal;i++)
        {
            while(cal%i==0)
            {
                if(a[i]>0)
                {
                    a[i]-=1;
                }else
                {
                    goto loop;
                }
                cal/=i;
            }
        }
        ans+=1;
    }
    loop: return ans;
}