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            posts - 195,  comments - 30,  trackbacks - 0
            #include <cstdio>
            #include 
            <string>

            int b[51][51][50], N;

            void add ( int i, int j )
            {
                
            int k;
                
            for ( k = 0; k < 50; k ++ )
                    b[i][j][k] 
            = b[i - 1][j - 1][k] + b[i - 1][j][k] * j;
                
            int c = 0, t;
                
            for ( k = 0; k < 50; k ++ )
                
            {
                    t 
            = b[i][j][k] + c;
                    c 
            = t / 10;
                    b[i][j][k] 
            = t % 10;
                }

            }


            void dp ()
            {
                memset ( b, 
            0x00sizeof ( b ) );
                
            int i, j;
                
            for ( i = 1; i <= 50; i ++ )
                
            {
                    b[i][
            1][0= b[i][i][0= 1;
                }

                
            for ( i = 3; i <= 50; i ++ )
                
            {
                    
            for ( j = 2; j < i; j ++ )
                    
            {
                        add ( i, j );
                    }

                }

            }


            void print ( int i, int j )
            {
                
            int k;
                
            for ( k = 49; k >= 0; k -- )
                    
            if ( b[i][j][k] )
                        
            break;
                
            if ( k == -1 )
                    printf ( 
            "0" );
                
            for ( ; k >= 0; k -- )
                    printf ( 
            "%d", b[i][j][k] );
                printf ( 
            " " );
            }


            void print ( int n )
            {
                printf ( 
            "%d ", n );
                
            int i, j, k;
                
            int ans[50];
                memset ( ans, 
            0sizeof ( ans ) );
                
            for ( i = 1; i <= n; i ++ )
                
            {
                    
            for ( j = 0; j < 50; j ++ )
                    
            {
                        ans[j] 
            += b[n][i][j];
                    }

                }

                
            int t, c = 0;
                
            for ( k = 0; k < 50; k ++ )
                
            {
                    t 
            = ans[k] + c;
                    c 
            = t / 10;
                    ans[k] 
            = t % 10;
                }

                
            for ( k = 49; k >= 0; k -- )
                    
            if ( ans[k] )
                        
            break;
                
            if ( k == -1 )
                    printf ( 
            "0" );
                
            for ( ; k >= 0; k -- )
                    printf ( 
            "%d", ans[k] );
                printf ( 
            " " );
            }


            int main ()
            {
                
            //freopen ( "in.txt", "r", stdin );
                dp ();
                
            //print ( 5, 2 );
                while ( scanf ( "%d"&N ) && N )
                
            {
                    print ( N );
                }

                
            return 0;
            }

            Rhyme Schemes
            Status In/Out TIME Limit MEMORY Limit Submit Times Solved Users JUDGE TYPE
            stdin/stdout 3s 8192K 97 55 Special Test

            The rhyme scheme for a poem (or stanza of a longer poem) tells which lines of the poem rhyme with which other lines. For example, a limerick such as

            
            If computers that you build are quantum
            Then spies of all factions will want 'em
            Our codes will all fail
            And they'll read our email
            `Til we've crypto that's quantum and daunt 'em
            Jennifer and Peter Short(http://www.research.att.com/~shor/notapoet.html)
            
            Has a rhyme scheme of aabba, indicating that the first, second and fifth lines rhyme and the third and fourth lines rhyme.

            For a poem or stanza of four lines, there are 15 possible rhyme schemes: aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, and abcd.

            Write a program to compute the number of rhyme schemes for a poem or stanza of N lines where N is an input value.

            Input

            Input will consist of a sequence of integers N, one per line, ending with a 0 (zero) to indicate the end of the data. N is the number of lines in a poem.

            Output

            For each input integer N, your program should output the value of N, followed by a space, followed by the number of rhyme schemes for a poem with N lines as a decimal integer with at least 12 correct significant digits (use double precision floating point for your computations).

            Sample Input

            1
            2
            3
            4
            20
            30
            10
            0
            

            Sample Output

            1 1
            2 2
            3 5
            4 15
            20 51724158235372
            30 846749014511809120000000
            10 115975
            


             
            排列組合的題目,大致上從小到大順推即可。設b[i][j]中i表示字符串長度,j表示字符串中用到的字母個數,不難推出b[i][j] = b[i-1][j-1] + b[i - 1][j] * j。
            而我自己的思路一直是想根據最后的那一個字母來推,推不出來

            posted on 2009-05-12 11:42 luis 閱讀(350) 評論(0)  編輯 收藏 引用 所屬分類: 組合數學
            <2009年5月>
            262728293012
            3456789
            10111213141516
            17181920212223
            24252627282930
            31123456

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