原題鏈接：HDU’s ACM 1042 N!

分析：這是一個顯然的大數問題，因爲N在[0,10000]，分析可知結果位數會在10^4數量級，若逐數位處理太慢，考慮多數位處理，但要注意N的不同範圍取值時，數組每個元素所能容納的最大值。另外，可以考慮存儲已經處理的數據，來提升性能。

AC Code:

#include <stdio.h>
#include <string.h>

#define MAXN 10000 + 2000

int MAX_NUM_PER_NUM = 100000;
int res[MAXN];

int main()
{
	int N;
	int i, j, len;
	while(scanf("%d", &N) == 1){
		memset(res, 0, sizeof(int)*MAXN);
		i = j = len = 0;
		res[0] = 1;
		if(N == 10000)
			MAX_NUM_PER_NUM = 10000;
		else
			MAX_NUM_PER_NUM = 100000;
		
		for(i = 1;i<=N;++i){
			for(j = 0;j<=len;++j)
				res[j] *= i;
			for(j = 0;j<=len;++j){
				res[j+1] += res[j] / MAX_NUM_PER_NUM;
				res[j] %= MAX_NUM_PER_NUM;
			}
			len = res[len+1]? len+1 : len;
		}
		printf("%d", res[len--]);
		if(N == 10000)
			while(len>=0)
				printf("%04d", res[len--]);
		else
			while(len>=0)
				printf("%05d", res[len--]);
		printf("\n");
	}
	return 0;
}

WC Code(待糾正)：

#include <stdio.h>
#include <string.h>

#define MAXN 10000 + 2000

int MAX_NUM_PER_NUM = 100000;
int res[MAXN];

int main()
{
	int N;
	int i, j, len;
	int tmp;
	int digitNum; // 標識數組每個元素可以容納的最大數字的數位,供輸出使用
	while(scanf("%d", &N) == 1){
		memset(res, 0, sizeof(int)*MAXN);
		i = j = len = 0;
		res[0] = 1;

		tmp = N/10;
		digitNum = 9;
		MAX_NUM_PER_NUM = 1000000000; // MAX_NUM_PER_NUM = 10^9
		while(tmp){
			MAX_NUM_PER_NUM /= 10;
			tmp /= 10;
			digitNum--;
		}
		for(i = 1;i<=N;++i){
			for(j = 0;j<=len;++j)
				res[j] *= i;
			for(j = 0;j<=len;++j){
				res[j+1] += res[j] / MAX_NUM_PER_NUM;
				res[j] %= MAX_NUM_PER_NUM;
			}
			len = res[len+1]? len+1 : len;
		}
		printf("%d", res[len--]);
		while(len>=0)
			printf("%0*d", digitNum, res[len--]);
		printf("\n");
	}
	return 0;
}