Posted on 2008-09-09 17:16
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代码如诗－－ACM
1 //1715 Accepted 204K 0MS C++ 1314B PKU
2
3 /*
4 1. 首先确定字符串的长度xlen
5 2. 对每一位进行枚举--不断缩小范围
6 */
7 #include <stdio.h>
8 #include <string.h>
9 #include <stdlib.h>
10
11 int inn ;
12 int num ;
13
14 char out[10] ;
15 int numout[10] ;
16
17 int A( int n, int m )
18 {
19 int reval = 1 ;
20
21 for( int i=0; i<m; i++ ) reval *= (n-i) ;
22
23 return reval ;
24 }
25
26 int main()
27 {
28 while( scanf( "%d", &inn ) != EOF )
29 {
30 int xlen = 888 ;
31 for( xlen=8; xlen>=1; xlen-- )
32 {
33 num = A( 15, xlen-1 ) * 15 ;
34 if( inn > num ) inn = inn - num ;
35 else break ;
36 }//确定长度为xlen
37
38 //printf( "xlen == %d\n", xlen ) ;
39
40 int first ;
41 for( first=15; first>=1; first-- )
42 {
43 num = A( 15, xlen-1 ) ;
44 if( inn > num) inn -= num ;
45 else
46 {
47 numout[1] = first ; break ;
48 }
49 }
50
51 for( int cxlen=2; cxlen<=xlen; cxlen++ )
52 {
53 for( int curposi=15; curposi>=0; curposi-- )
54 {
55 bool contin = false ;
56 for( int i=1; i<cxlen; i++ )
57 {
58 if( curposi == numout[i] ) { contin = true ; break ;}
59 }
60 if( contin ) continue ;
61
62 num = A( 16-cxlen, xlen-cxlen ) ;
63 if( inn > num ) inn -= num ;
64 else
65 {
66 numout[cxlen] = curposi ; break ;
67 }
68 }
69 }
70
71 for( int i=1; i<=xlen; i++ )
72 {
73 if( numout[i]>=0 && numout[i]<=9 )
74 printf( "%d", numout[i] ) ;
75 else
76 printf( "%c", numout[i]-10+'A' ) ;
77 }
78 printf( "\n" ) ;
79
80 }//while
81
82 return 0 ;
83 }