题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1042

题目大意：求n！， n 的上限是10000。

解题思路：高精度乘法 ， 因为数据量比较大， 所以得用到缩进，但因为是乘法，缩进只能在10^5， 不然在乘的过程中就超过int型。然后数值位数大概在8000位以下。

#include <iostream>

#include <string.h>

using namespace std;

const int MAXN = 100000;

const int N = 8000;

struct bign {

    int len;

    int s[N];

    bign() {

        this -> len = 1;

        memset(s, 0, sizeof(s));

    }

    bign (int number) {*this = number;}

    bign (const char* number) {*this = number;}

    bign change(bign cur) {

        bign now;

        now = cur;

        for (int i = 0; i < cur.len; i++)

            now.s[i] = cur.s[cur.len - i - 1];

        return now;

    }

    void delZore() {	// 删除前导0.

        bign now = change(*this);

        while (now.s[now.len - 1] == 0 && now.len > 1) {

            now.len--;

        }

        *this = change(now);

    }

    void put() {    // 输出数值。

        delZore();

        printf("%d", s[0]);

        for (int i = 1; i < len; i++)

            printf("%05d", s[i]);

    }

    bign operator = (const char *number) {

        memset(s, 0, sizeof(s));

        int dist = strlen(number);

        int k = dist % 8;

        for (int i = 0; i < k; i++)

            s[0] = s[0] * 10 + number[i] - '0';

	int cnt = 0;

	for (int i = k; i < dist; i++, cnt++)

	    s[cnt / 8 + 1] = s[cnt / 8 + 1] * 10 + number[i] - '0';

	len = cnt / 8 + 1;

	return *this;

    }

    bign operator = (int number) {

	char string[N];

	sprintf(string, "%d", number);

	*this = string;

	return *this;

    }

    bign operator * (const bign &cur){

	bign sum, a, b;

	sum.len = 0;

	a = a.change(*this);

	b = b.change(cur);

	for (int i = 0; i < a.len; i++){

	    int g = 0;  

	    for (int j = 0; j < b.len; j++){

		int x = a.s[i] * b.s[j] + g + sum.s[i + j];

		sum.s[i + j] = x % MAXN;

		g = x / MAXN;

	    }

	    sum.len = i + b.len;  

	    while (g){

		sum.s[sum.len++] = g % MAXN;

		g = g / MAXN;

	    }

	}

	return sum.change(sum);

    }

};

int main() {

    int n;

    while (scanf("%d", &n) == 1) {

	bign sum = 1;

	for (int i = 2; i <= n; i++) {

	    bign now = i;

	    sum = sum * now;

	}

	sum.put();

	printf("\n");

    }

    return 0;

}