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 = 5002;
const int mod = 1e6;

int b[M + 3][2800], len[M + 3];

inline void init(int n = M) {
    b[1][0] = 1;    len[1] = 1;
    for(int i = 2; i <= n; ++i) {
        int x = len[i - 1];
        int carry = 0;
        LL tmp;
        for(int j = 0; j < x; ++j) {
            tmp = carry + (LL)b[i - 1][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 / 100000 % 10 + '0');
    putchar(x / 10000 % 10 + '0');
    putchar(x / 1000 % 10 + '0');
    putchar(x / 100 % 10 + '0');
    putchar(x / 10 % 10 + '0');
    putchar(x % 10 + '0');
}

inline void output(const int &n) {
    int i = len[n] - 1;
    const int &x = b[n][i];
    printf("%d",x);

    for(--i; i >= 0; --i)
        print(b[n][i]);
    puts("");
}

int main() {
    int m;
    init();
    while(~scanf("%d",&m))
        output(m);
    return 0;
}

　　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 = 5002;

int b[2][18000];

inline void solve(const int &n) {
    b[1][0] = 1;
    int len = 1;
    for(int i = 2; i <= n; ++i) {
        int carry = 0, tmp;
        for(int j = 0; j < len; ++j) {
            tmp = carry + b[!(i & 1)][j] * i;
            b[i & 1][j] = tmp % 10;
            carry = tmp / 10;
        }
        while(carry) {
            b[i & 1][len++] = carry % 10;
            carry /= 10;
        }
    }
    for(int j = len - 1; j >= 0; --j)
        putchar(b[n & 1][j] + '0');
    puts("");
}

template <typename T>
inline bool read(T &x) {
    x = 0;
    char ch = getchar();
    while(!isdigit(ch) && ch != EOF)    ch = getchar();
    if(ch == EOF)   return 0;
    while(isdigit(ch)) {
        x = x * 10 + (ch - '0');
        ch = getchar();
    }
    return 1;
}

int main() {
    int m;
    while(read(m))
        solve(m);
    return 0;
}