UVA

n!=x*b^y,

当x为正整数时,最大的y就是n!末尾0的个数了,

把n,b分别拆成素因子相乘的形式:

比如,

n=5,b=16

n=5,b=2^4,

非常明显,末尾0的个数为0

10进制时,n!=a*10^x

b进制时,n!=c*b^y

非常明显,n!的位数就是最大的x+1

这里计算我用了log,精度设置为1e-9

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<cmath>
using namespace std;
const int inf=(1<<31)-1;
const double eps=1e-9;
vector<int>prime;
void maketable()
{
    int i,j,n=800;
    bool iscp[810];
    memset(iscp,0,sizeof(iscp));
    for(i=2;i<=n;i++)
    {
        if(!iscp[i])
        {
            prime.push_back(i);
            for(j=i+i;j<=n;j+=i)
                iscp[j]=1;
        }
    }
}
map<int,int>fn;
map<int,int>fb;
map<int,int>::iterator it;
void debug()
{
    cout<<"***************"<<endl;
    for(it=fn.begin();it!=fn.end();it++)
        cout<<it->first<<"^"<<it->second<<endl;
    cout<<"***************"<<endl;
     for(it=fb.begin();it!=fb.end();it++)
        cout<<it->first<<"^"<<it->second<<endl;
     cout<<"***************"<<endl;
}
int main()
{
    //freopen("in","r",stdin);
    //freopen("out","w",stdout);
    maketable();
    int i,j,k,n,b,dg,m,num_zero;
    double x;
    while(cin>>n>>b)
    {
        fn.clear();
        fb.clear();
        x=0;
        for(i=2;i<=n;i++)
            x+=log10(double(i));
        dg=int(x/log10(double(b))+eps)+1;
        m=prime.size();
        for(i=2;i<=n;i++)
        {
            k=i;
            for(j=0;j<m&&k>=prime[j];j++)
            {
                while(k%prime[j]==0&&k>=prime[j])
                {
                    fn[prime[j]]++;
                    k/=prime[j];
                }
            }
        }
        for(i=0;i<m&&b>=prime[i];i++)
        {
            while(b%prime[i]==0&&b>=prime[i])
            {
                fb[prime[i]]++;
                b/=prime[i];
            }
        }
        //debug();
        num_zero=inf;
        for(it=fb.begin();it!=fb.end();it++)
            num_zero=min(num_zero,fn[it->first]/it->second);
        cout<<num_zero<<" "<<dg<<endl;
    }
    return 0;
}

Problem G

How many zeros and how many digits?

Input: standard input

Output: standard output

Given a decimal integer number you willhave to find out how many trailing zeros will be there in its factorial in a given number system and alsoyou will have to find how many digits will its factorial have in a given number system? You can assume that fora 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 linewill 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 ina single line how many trailing zeros will the factorial of that numberhave in the given number system and also how many digits will the factorial of thatnumber have in that given number system. Separate these two numbers with a single space. You can be surethat the number of trailing zeros or the number of digits will not be greaterthan 2^31-1

Sample Input:

2 10
5 16
5 10

 

Sample Output:

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