【思路】：大数基本都是这思路，，采用数组或者字符串，每个数采用倒序的方式从头开始存储。每次进位进到下一位上。
【AC代码】：两个数组来回颠倒。
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <iomanip>using namespace std;#define MAX 3000int cal(int *s, int len, int n, int *d){int i = 0, temp = 0;for (i = 0; i < len; i++){temp = s[i] * n + temp;d[i] = temp%10;temp = temp/10;}while (temp){d[i++] = temp%10;temp = temp/10;}return i;}void output(int *s, int len){int i = 0;while (len){cout << s[–len];}}int main(){//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);int n =0, i = 0, j = 1, k = 0;int s1[MAX]={1}, s2[MAX];cin >> n;for (i = 1; i <= n; i++){if (j > 0){k = cal(s1, j, i, s2);j = 0;}else{j = cal(s2, k, i, s1);k = 0;}}if (0 == n%2)output(s1, j);elseoutput(s2, k);}【AC代码2】：一个数组。
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <iomanip>using namespace std;#define MAX 3000int cal(int *s, int len, int n){int i = 0, temp = 0;for (i = 0; i < len; i++){temp = s[i] * n + temp;s[i] = temp%10;temp = temp/10;}while (temp){s[i++] = temp%10;temp = temp/10;}return i;}void output(int *s, int len){int i = 0;while (len){cout << s[–len];}}int main(){//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);int n =0, i = 0, len = 1;int s1[MAX]={1};cin >> n;for (i = 1; i <= n; i++){len = cal(s1, len, i);}output(s1, len);}
人生就是一场旅行，不在乎目的地，
