joj 1978

<转帖> joj 1978

#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, 0x00, sizeof ( 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, 0, sizeof ( 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

 
排列组合的题目，大致上从小到大顺推即可。设b[i][j]中i表示字符串长度，j表示字符串中用到的字母个数，不难推出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据最后的那一个字母来推，推不出来