hdu 1027 Ignatius and the Princess II 康托展开式的逆过程

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康托展开式的逆过程，也可以用库函数写

//康托展开式（不懂百度百科，写的很详细）这题是一个逆过程//8位的数一共有40000多种排序远大于题目要求的10000；//所以八位以前的数不变。#include<iostream>using namespace std;__int64 fac[]={1,1,2,6,24,120,720,5040,40320,362880,362880};int ch[1000];int n, num;void ss(){int i, j, k, mode, nn=n;memset(ch, 0, sizeof(ch));if(n > 8)//超过8位的前面1~n-8不变，后面排序{for( i=0; i < n-8; i++ )ch[i] = i+1;nn = 8;}num--;for( i=nn-1; i > 0; i-- ){mode = num % fac[i];num /= fac[i];int temp=0;for( j=1; j <= n; j++ )//遍历在这个位子以前哪些数是用过的{for( k=0; k < n-i-1; k++ ){if(ch[k] == j)break;}if(k == n-i-1)temp++;if(temp == num+1){ch[n-i-1]=j;break;}}num = mode;}for(i=1; i <= n; i++ )//寻找最后一位数{for(j=0; j < n; j++ ){if(ch[j] == i)break;}if(j == n){ch[n-1] = i;break;}}}int main(){while(scanf("%d%d", &n, &num) != EOF ){ss();bool p = false;for(int i=0; i < n; i++ ){if(p)printf(" ");p = true;printf("%d", ch[i]);}printf("\n");}return 0;}

库函数next_permutstion

//库函数next_permutaion(a, a+n)#include<iostream>#include<algorithm>using namespace std;int main(){int n, m, i, j, a[1005];while(scanf("%d%d", &n, &m) != EOF){for( i=0; i < n; i++ )a[i] = i+1;while(--m) { next_permutation(a, a+n); }bool p = false;for( i=0; i < n; i++ ){if(p)printf(" ");p = true;printf("%d", a[i]);}printf("\n");}return 0;}