LightOJ 1060 - nth Permutation （逆康托展开的思想）

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题意:

求n<=20长度串的所有排列的按照字典序的第k个序列

分析:

总排列数f=n!/(cnta!∗cntb!∗⋯∗cntz!)
恢复序列的思想类似于逆康托展开，点击这里
这个题也可以用状压dp来做−−弱表示不会

代码:

////  Created by TaoSama on 2015-11-20//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n, cnt[26];LL k, fact[25] = {1};char s[25];int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    for(int i = 1; i <= 20; ++i) fact[i] = fact[i - 1] * i;    while(t--) {        scanf("%s%lld", s + 1, &k);        n = strlen(s + 1);        memset(cnt, 0, sizeof cnt);        for(int i = 1; i <= n; ++i) cnt[s[i] - 'a']++;        printf("Case %d: ", ++kase);        LL maxk = fact[n];        for(int i = 0; i < 26; ++i) maxk /= fact[cnt[i]];        if(k > maxk) {puts("Impossible"); continue;}        for(int i = 1; i <= n; ++i) {            for(int j = 0; j < 26; ++j) {                if(!cnt[j]) continue;                --cnt[j];                LL tmp = fact[n - i];                for(int k = 0; k < 26; ++k)                    tmp /= fact[cnt[k]];                if(k <= tmp) {                    putchar('a' + j);                    break;                }                ++cnt[j];                k -= tmp;            }        }        puts("");    }    return 0;}

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