HDU 1027 Ignatius and the Princess II（全排列）

题意： 对于给定的 1-n的序列，输出第m个全排列
思路： 注意到m最多只有10000，故最多只需要改变末尾的k个，对末尾的k个进行全排列，然后顺序输出

//#include <iostream>//#include <cstdio>//#include <string>//#include <cstring>//#include <string.h>//#include <cmath>//#include <algorithm>//#include <cctype>//#include <cstdlib>//#include <ctime>//#include <queue>#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int, int> pii;ll dp[50];int main(){    int n, m;    while (scanf("%d%d", &n, &m) != EOF)    {        int all = 1; int p = 1;        while (all < m)++p,all *= p;        vector<int>v;        for (int i = n - p + 1; i <= n; i++)v.push_back(i);        do        {            m--;            if (m == 0)break;        } while (next_permutation(v.begin(), v.end()));        int flag = 0;        for (int i = 1; i <= n - p; i++)        {            if (!flag)flag = 1;            else printf(" ");            printf("%d", i);        }        for (int i = 0; i < v.size(); i++)        {            if (!flag)flag = 1;            else printf(" ");            printf("%d", v[i]);        }        printf("\n");    }}