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1060 - nth Permutation

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Time Limit: 2 second(s)Memory Limit: 32 MB

Given a string of characters, we can permute the individual characters to make new strings. At first we order the string into alphabetical order. Then we start permuting it.

For example the string 'abba' gives rise to the following 6 distinct permutations in alphabetical order.

aabb 1

abab 2

abba 3

baab 4

baba 5

bbaa 6

Given a string, you have to find the n^th permutation for that string. For the above case 'aabb' is the 1^st and 'baab' is the 4^th permutation.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a non empty string of lowercase letters with length no more than 20 and an integer n (0 < n < 2³¹).

Output

For each case, output the case number and the n^th permutation. If the n^th permutation doesn't exist print 'Impossible'.

Sample Input

Output for Sample Input

3

aabb 1

aabb 6

aabb 7

Case 1: aabb

Case 2: bbaa

Case 3: Impossible

题意：给一个小写字母组成的字符串，求这个字符串字典序排序后的第k个字符串

#include <iostream>#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1e6+7;const int inf = 0x3f3f3f3f;const int mod = 10007;char str[30];int num[30];LL c[30][30];LL judge(int n){    LL ans=1;    for(int i=0; i<26; i++)    {        if(!num[i]) continue;        ans*=c[n][num[i]];        n-=num[i];    }    return ans;}int main(){    memset(c,0,sizeof(c));    for(int i=0; i<=27; i++) c[i][0]=c[i][i]=1;    for(int i=2; i<=27; i++)        for(int j=1; j<i; j++)            c[i][j]=c[i-1][j]+c[i-1][j-1];    int ncase=1, t;    LL n;    scanf("%d", &t);    while(t--)    {        scanf("%s %lld",str, &n);        printf("Case %d: ",ncase++);        memset(num,0,sizeof(num));        int len=strlen(str);        for(int i=0; str[i]; i++) num[str[i]-'a']++;        if(judge(len)<n) puts("Impossible");        else        {            for(int i=1; i<=len; i++)            {                for(int j=0; j<26; j++)                {                    if(!num[j]) continue;                    num[j]-=1;                    LL tmp=judge(len-i);                    if(tmp>=n)                    {                        printf("%c",j+'a');                        break;                    }                    num[j]+=1;                    n-=tmp;                }            }            cout<<endl;        }    }    return 0;}