Lexicography
Time Limit:1000MSMemory Limit:131072KB64bit IO Format:%lld & %llu
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Description
An anagram of a string is any string that can be formed using the same letters as the original. (We consider the original string an anagram of itself as well.) For example, the string ACM has the following 6 anagrams, as given in alphabetical order:
ACMAMCCAMCMAMACMCAAs another example, the string ICPC has the following 12 anagrams (in alphabetical order):
CCIPCCPICICPCIPCCPCICPICICCPICPCIPCCPCCIPCICPICCGiven a string and a rank K, you are to determine the Kth such anagram according to alphabetical order.
Input
Each test case will be designated on a single line containing the original word followed by the desired rank K. Words will use uppercase letters (i.e., A through Z) and will have length at most 16. The value of K will be in the range from 1 to the number of distinct anagrams of the given word. A line of the form "# 0" designates the end of the input.
Output
For each test, display the Kth anagram of the original string.
Sample Input
ACM 5ICPC 12REGION 274# 0
Sample Output
MACPICCIGNORE
Hint
The value of K could be almost 245 in the largest tests, so you should use type long in Java, or type long long in C++ to store K.
题意：给定一个串和k，求这个串字典序第k的序列。
代码；
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b) for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i–)#define FRL(i,a,b) for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define DBGpf("Hi\n")typedef long long ll;using namespace std;ll fac[20],k;char str[30];int num[30];ll getnum(int l){ll s=fac[l];for (int i=0;i<26;i++)s/=fac[num[i]];return s;}void solve(int len){for (int i=0;i<len;i++){for (int j=0;j<26;j++){if (num[j]){num[j]–;ll x=getnum(len-i-1);num[j]++;if (x>=k){num[j]–;printf("%c",j+'A');break;}else k-=x;}}}printf("\n");}int main(){// freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);int i,j;fac[0]=1;for (i=1;i<=17;i++)fac[i]=fac[i-1]*i;while (scanf("%s %lld",str,&k)){if (k==0) break;int len=strlen(str);mem(num,0);for (i=0;i<len;i++)num[str[i]-'A']++;solve(len);}return 0;}
，黑夜下，撕开那张面具尽是怠倦的容颜，
