求n!的位数以及求n!具体的值（C or C++）

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首先我们先求n!位数
可以将n!表示成10的次幂,即n!=10^M(10的M次方)则不小于M的最小整数就是 n!的位数，对该式两边取对数，有 M =log10^n!
即：
M = log10^1+log10^2+log10^3…+log10^n
循环求和,就能算得M值，该M是n!的精确位数。

#include<iostream>#include<cmath>using namespace std;int main(){    int n;    int i;    double d;    while (cin>>n)    {        d=0;        for (i=1;i<=n;i++)        {            d+=(double)log10(i);        }        cout<<(int)d+1<<endl; } return 0;}

接下来，求n!的具体值
具体题目原型来自HDOJ
题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1042

C++ Version:

#include <iostream>#include <cstring>/* 一个数组元素表示 4 个十进制位，即数组是万进制的 */#define BIG_RADIX 10000#define RADIX_LEN 4/* 10000! 有 35660 位 */#define MAX_LEN (35660/RADIX_LEN + 1)using namespace std;int x[MAX_LEN + 1];void Big_Print(){    int start_output = 0;//用于跳过多余的0    for (int i=MAX_LEN;i>= 0;--i)    {        if (start_output == 1){//如果多余的0已经跳过，则输出            cout<<x[i];        }        else if (x[i] > 0){//表示多余的0已经跳过            cout<<x[i];            start_output = 1;        }    }    if (start_output == 0)//如果全为0        cout<<"0";}void Big_Multiple(int y){    int carry = 0;//保存进位    int tmp;    for (int i=0;i<MAX_LEN;++i)    {        tmp = x[i] * y + carry;        x[i] = tmp % BIG_RADIX;        carry = tmp / BIG_RADIX;    }}void Big_Factorial(int N){    memset(x, 0, sizeof(x));    x[0] = 1;    for (int i=2; i<=N;++i){        Big_Multiple(i);    }}int main(void){    int N;    while (cin>>N)    {        Big_Factorial(N);        Big_Print();        cout<<endl;    }    return 0;}

C Version:

#include <stdio.h>  #include <string.h>  /* 一个数组元素表示 4 个十进制位，即数组是万进制的 */  #define BIG_RADIX 10000  #define RADIX_LEN 4  /* 10000! 有 35660 位 */  #define MAX_LEN (35660/RADIX_LEN + 1)  int x[MAX_LEN + 1];  void Big_Print(){      int i;      int start_output = 0;//用于跳过多余的0      for (i = MAX_LEN; i >= 0; --i){          if (start_output == 1){//如果多余的0已经跳过，则输出              printf("%04d", x[i]);          }          else if (x[i] > 0){//表示多余的0已经跳过              printf("%d", x[i]);              start_output = 1;          }      }      if (start_output == 0)//如果全为0          printf("0");  }  void Big_Multiple(int y){      int i;      int carry = 0;//保存进位      int tmp;      for (i = 0; i < MAX_LEN; ++i){          tmp = x[i] * y + carry;          x[i] = tmp % BIG_RADIX;          carry = tmp / BIG_RADIX;      }  }  void Big_Factorial(int N){      int i;      memset(x, 0, sizeof(x));      x[0] = 1;      for (i = 2; i <= N; ++i){          Big_Multiple(i);      }  }  int main(void){      int N;      while (scanf("%d", &N) != EOF){          Big_Factorial(N);          Big_Print();          printf("\n");      }  }

Acknowledgements:
liushaobo:http://blog.csdn.net/jdplus/article/details/20376203

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