HDU 1042 N!（高精度乘）

Problem Description

Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

 

Input

One N in one line, process to the end of file.

 

Output

For each N, output N! in one line.

 

题目大意：求N的阶乘。

思路：用高精度，内存存不下这么多只能每次都重新算了……

代码（3093MS）：

 //模板测试
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 using namespace std;

 const int MAXN = ;

 struct bign {
     int len, s[MAXN];

     bign () {
         memset(s, , sizeof(s));
         len = ;
     }
     bign (int num) { *this = num; }
     bign (const char *num) { *this = num; }

     void clear() {
         memset(s, , sizeof(s));
         len = ;
     }

     bign operator = (const int num) {//数字
         char s[MAXN];
         sprintf(s, "%d", num);
         *this = s;
         return *this;
     }
     bign operator = (const char *num) {//字符串
         for(int i = ; num[i] == ''; num++) ;  //去前导0
         if(*num == ) --num;
         len = strlen(num);
         for(int i = ; i < len; ++i) s[i] = num[len-i-] - '';
         return *this;
     }

     bign operator + (const bign &b) const {
         bign c;
         c.len = ;
         for(int i = , g = ; g || i < max(len, b.len); ++i) {
             int x = g;
             if(i < len) x += s[i];
             if(i < b.len) x += b.s[i];
             c.s[c.len++] = x % ;
             g = x / ;
         }
         return c;
     }

     bign operator += (const bign &b) {
         *this = *this + b;
         return *this;
     }

     void clean() {
         while(len >  && !s[len-]) len--;
     }

     bign operator * (const bign &b) {
         bign c;
         c.len = len + b.len;
         for(int i = ; i < len; ++i) {
             for(int j = ; j < b.len; ++j) {
                 c.s[i+j] += s[i] * b.s[j];
             }
         }
         for(int i = ; i < c.len; ++i) {
             c.s[i+] += c.s[i]/;
             c.s[i] %= ;

         }
         c.clean();
         return c;
     }
     bign operator *= (const bign &b) {
         *this = *this * b;
         return *this;
     }

     bign operator *= (const int &b) {//使用前要保证>len的位置都是空的
         for(int i = ; i < len; ++i) s[i] *= b;
         for(int i = ; i < len; ++i) {
             s[i + ] += s[i] / ;
             s[i] %= ;
         }
         while(s[len]) {
             s[len + ] += s[len] / ;
             s[len] %= ;
             ++len;
         }
         return *this;
     }

     bign operator - (const bign &b) {
         bign c;
         c.len = ;
         for(int i = , g = ; i < len; ++i) {
             int x = s[i] - g;
             if(i < b.len) x -= b.s[i];
             if(x >= ) g = ;
             else {
                 g = ;
                 x += ;
             }
             c.s[c.len++] = x;
         }
         c.clean();
         return c;
     }
     bign operator -= (const bign &b) {
         *this = *this - b;
         return *this;
     }

     bign operator / (const bign &b) {
         bign c, f = ;
         for(int i = len - ; i >= ; i--) {
             f *= ;
             f.s[] = s[i];
             while(f >= b) {
                 f -= b;
                 c.s[i]++;
             }
         }
         c.len = len;
         c.clean();
         return c;
     }
     bign operator /= (const bign &b) {
         *this  = *this / b;
         return *this;
     }

     bign operator % (const bign &b) {
         bign r = *this / b;
         r = *this - r*b;
         return r;
     }
     bign operator %= (const bign &b) {
         *this = *this % b;
         return *this;
     }

     bool operator < (const bign &b) {
         if(len != b.len) return len < b.len;
         for(int i = len-; i >= ; i--) {
             if(s[i] != b.s[i]) return s[i] < b.s[i];
         }
         return false;
     }

     bool operator > (const bign &b) {
         if(len != b.len) return len > b.len;
         for(int i = len-; i >= ; i--) {
             if(s[i] != b.s[i]) return s[i] > b.s[i];
         }
         return false;
     }

     bool operator == (const bign &b) {
         return !(*this > b) && !(*this < b);
     }

     bool operator != (const bign &b) {
         return !(*this == b);
     }

     bool operator <= (const bign &b) {
         return *this < b || *this == b;
     }

     bool operator >= (const bign &b) {
         return *this > b || *this == b;
     }

     string str() const {
         string res = "";
         for(int i = ; i < len; ++i) res = char(s[i]+'') + res;
         return res;
     }
 };

 istream& operator >> (istream &in, bign &x) {
     string s;
     in >> s;
     x = s.c_str();
     return in;
 }

 ostream& operator << (ostream &out, const bign &x) {
     out << x.str();
     return out;
 }

 bign ans;

 void solve(int n) {
     ans.clear();
     ans.len = ans.s[] = ;
     for(int i = ; i <= n; ++i) ans *= i;
     cout<<ans<<endl;
 }

 int main() {
     int n;
     while(scanf("%d", &n)!=EOF) {
         //cout<<f[n]<<endl;
         solve(n);
     }
     return ;
 }