nyoj 28 大数阶乘

　　题目链接：nyoj 28

　　就是个简单的高精度，只是一开始我打表超内存了，然后用了各种技巧硬是把内存缩到了题目要求以下（5w+kb），感觉挺爽的，代码如下：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 typedef long long LL;
 const int M = ;
 const int mod = 1e6;

 int b[M + ][], len[M + ];

 inline void init(int n = M) {
     b[][] = ;    len[] = ;
     for(int i = ; i <= n; ++i) {
         int x = len[i - ];
         int carry = ;
         LL tmp;
         for(int j = ; j < x; ++j) {
             tmp = carry + (LL)b[i - ][j] * i;
             b[i][j] = tmp % mod;
             carry = tmp / mod;
         }
         while(carry) {
             b[i][x++] = carry % mod;
             carry /= mod;
         }
         len[i] = x;
     }
 }

 inline void print(const int &x) {
     putchar(x /  %  + '');
     putchar(x /  %  + '');
     putchar(x /  %  + '');
     putchar(x /  %  + '');
     putchar(x /  %  + '');
     putchar(x %  + '');
 }

 inline void output(const int &n) {
     int i = len[n] - ;
     const int &x = b[n][i];
     printf("%d",x);

     for(--i; i >= ; --i)
         print(b[n][i]);
     puts("");
 }

 int main() {
     int m;
     init();
     while(~scanf("%d",&m))
         output(m);
     return ;
 }

　　出题人原意应该不是让我们打表，而是每读入一个数重新计算一个数……吧：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<cctype>
 const int M = ;

 int b[][];

 inline void solve(const int &n) {
     b[][] = ;
     int len = ;
     for(int i = ; i <= n; ++i) {
         int carry = , tmp;
         for(int j = ; j < len; ++j) {
             tmp = carry + b[!(i & )][j] * i;
             b[i & ][j] = tmp % ;
             carry = tmp / ;
         }
         while(carry) {
             b[i & ][len++] = carry % ;
             carry /= ;
         }
     }
     for(int j = len - ; j >= ; --j)
         putchar(b[n & ][j] + '');
     puts("");
 }

 template <typename T>
 inline bool read(T &x) {
     x = ;
     char ch = getchar();
     while(!isdigit(ch) && ch != EOF)    ch = getchar();
     if(ch == EOF)   return ;
     while(isdigit(ch)) {
         x = x *  + (ch - '');
         ch = getchar();
     }
     return ;
 }

 int main() {
     int m;
     while(read(m))
         solve(m);
     return ;
 }