UVa 10061: How many zero's and how many digits?
2013-07-26 18:31
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这题有点复杂,题目要求十进制数N!在B进制下表示后末尾的0个数和总的位数。
分析如下:
1. 求末尾0的个数:
思路是将N!分解为素数的积,记录下<=B的素数因子的个数(小于等于B的原因代码中解释了)。然后不断地从N!的因子中提取因子,使之刚好能乘积为B,则这几个因子可使N!在B进制下末尾产生一个0,直到提取不出为止。(当然,写代码的思路是不断分解B,当不能找到N!中小于等于B的因子用来分解B时结束)。记录下上述过程中B被分解了几次,即在末尾产生了几个0.
2. 求总的位数:
由于B进制下m位数所能表示的最大值在十进制下为B^m-1,故
设总位数为m,则 B^(m-1) -1 < N! <= B^m -1,即 B^(m-1) <= N! < B^m
取10为底的对数得到: (m-1)*log10(B) <= log10(N!) < m*log10(B)
故得到求总的位数m的方法。
我的解题代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
using namespace std;
int factor_count[1000]; //用来记录小于等于B的因子的个数
int N,B;
int ZeroNum()
{
//decompose N!
memset(factor_count,0,sizeof(factor_count));
for(int i=2; i<=N; i++)
{//decompose i
int tmp=i;
for(int j=2; j<=tmp && j<=B; j++) //只需要小于等于B的因子,因为大于B的因子一定不会成为B的因子
{
while(tmp%j==0)
{
factor_count[j]++;
tmp /= j;
}
}
}
int nZero=0;
while(1)
{//decompose B
int tmp=B;
for(int i=2; i<=tmp; i++)
{
while(tmp%i==0 && factor_count[i]>0)
{
factor_count[i]--;
tmp /= i;
}
}
if(tmp==1) nZero++;
else break;
}
return nZero;
}
int DigitNum()
{
//compute logB(N!)
double sum=0;
for(int i=2; i<=N; i++)
sum += log10(double(i));
sum /= log10(double(B));
//comput digits num
return floor(sum+1e-9) + 1;
}
int main()
{
while(cin >> N >> B)
{
cout << ZeroNum() << ' ' << DigitNum() << endl;
}
return 0;
}
分析如下:
1. 求末尾0的个数:
思路是将N!分解为素数的积,记录下<=B的素数因子的个数(小于等于B的原因代码中解释了)。然后不断地从N!的因子中提取因子,使之刚好能乘积为B,则这几个因子可使N!在B进制下末尾产生一个0,直到提取不出为止。(当然,写代码的思路是不断分解B,当不能找到N!中小于等于B的因子用来分解B时结束)。记录下上述过程中B被分解了几次,即在末尾产生了几个0.
2. 求总的位数:
由于B进制下m位数所能表示的最大值在十进制下为B^m-1,故
设总位数为m,则 B^(m-1) -1 < N! <= B^m -1,即 B^(m-1) <= N! < B^m
取10为底的对数得到: (m-1)*log10(B) <= log10(N!) < m*log10(B)
故得到求总的位数m的方法。
我的解题代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
using namespace std;
int factor_count[1000]; //用来记录小于等于B的因子的个数
int N,B;
int ZeroNum()
{
//decompose N!
memset(factor_count,0,sizeof(factor_count));
for(int i=2; i<=N; i++)
{//decompose i
int tmp=i;
for(int j=2; j<=tmp && j<=B; j++) //只需要小于等于B的因子,因为大于B的因子一定不会成为B的因子
{
while(tmp%j==0)
{
factor_count[j]++;
tmp /= j;
}
}
}
int nZero=0;
while(1)
{//decompose B
int tmp=B;
for(int i=2; i<=tmp; i++)
{
while(tmp%i==0 && factor_count[i]>0)
{
factor_count[i]--;
tmp /= i;
}
}
if(tmp==1) nZero++;
else break;
}
return nZero;
}
int DigitNum()
{
//compute logB(N!)
double sum=0;
for(int i=2; i<=N; i++)
sum += log10(double(i));
sum /= log10(double(B));
//comput digits num
return floor(sum+1e-9) + 1;
}
int main()
{
while(cin >> N >> B)
{
cout << ZeroNum() << ' ' << DigitNum() << endl;
}
return 0;
}
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