【模板】康拓展开与康拓逆展开

康拓展开与逆展开链接：https://www.cnblogs.com/Howe-Young/p/4348777.html(标准模板)

康拓展开：比当前位置小的数的个数*阶乘 

习题：http://nyoj.top/problem/139

---

#include <iostream>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <cstdio>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const ll maxn = 1e5+100;
const ll mod = 998244353;
const ld pi = acos(-1.0);
const ll inf = 1e18;

ll n,fac[15];
bool flag[15];

void getnum()
{
    fac[0] = 1;
	fac[1] = 1;
	for(ll i = 2; i <= 15; i++)
	{
		fac[i] = fac[i-1]*i;
	}
	return ;
}

int main()
{
	ios::sync_with_stdio(false);

	getnum();	// 获取阶乘结果 
	
	cin >> n;
	
	while(n--)
	{
		
		memset(flag,0,sizeof(flag));
		string s;
		cin >> s;
		
		ll ans = 0;
		
		for(ll i = 0; i < s.size()-1; i++)
		{
			ll num = 0;
			flag[ s[i]-'a' ] = 1;  //标记出现过 
			
			for(ll j = 0; j < s[i]-'a'; j++)  //枚举比s[i]小的字符 
			{
				if(flag[j] == 0)
					num++;    
			}
			
			ans += num*fac[ 12 - i - 1 ];  //个数*阶乘 
			
		}	
		cout << ans+1 << endl;	
	}
	
	return 0;
}

---

康拓逆展开：

---

习题：http://nyoj.top/problem/143

 注意阶乘数组 fac[0] = 1

---

#include <iostream>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <cstdio>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const ll maxn = 1e5+100;
const ll mod = 998244353;
const ld pi = acos(-1.0);
const ll inf = 1e18;

ll n,fac[15],T;
bool flag[15];

void getnum()
{
	fac[0] = 1;
	fac[1] = 1;
	for(ll i = 2; i <= 15; i++)
	{
		fac[i] = fac[i-1]*i;
	}
	return ;
}

int main()
{
	ios::sync_with_stdio(false);

	getnum();	// 获取阶乘结果 
	
	cin >> T;
	
	while(T--)
	{
		memset(flag,0,sizeof(flag));
		
		cin >> n;
		n--;
		
		ll len = 12;  //字符串长度 
		
		string ans;
		
		for(ll i = 1; i <= len; i++)
		{
			
			ll t = n/fac[len - i]; //代表有t个数字比当前数字小 
			n = n%fac[len-i];
			
			ll temp = 0,num = 0;  //temp表示用过的数，num表示没用过的 
			for(ll j = 0; j < len; j++)
			{
				if(num > t)  //找到t个没用过的数字就可以停止了 
					break;
					
				if(flag[j] == 1) // 用过的 
				{
					temp++;
				}
				else
				{
					num++;
				}
			}
			
			flag[temp+num-1] = 1;
			ans += char('a'+ (temp+num-1) );
			//cout << ans << endl;
		}
		
		cout << ans << endl;
			
	}
	return 0;
}