编辑本段证明

其实就是把A个红球，B个篮球分成n+1堆（除n堆外还有一堆就是没有放入盒中的）。
思路很重要，
我的思路是由于A球，B球不同色，可以看成两个独立事件，则结果为count(A)*count(B),
其中count(A)表示A的放法，
则利用隔板法(或者多重集的r组合)：count(A)=C(n+1+A-1,n+1-1   ),利用pascal公式c(n,m)=c(n-1,m)+c(n-1,m-1)来求该式;

还有一个注意点当且仅当n=20,a= 15 ,b=15时会超出long long 范围。所以特殊处理

点"+"展开
#include<iostream>
#include<cstdlib>
using namespace std;
  long long num[40][40];  
  long long fun(int n,int m)
{
   if(n<=0)
   return 0;
   if(num[n][m]!=0)
   return num[n][m];
   //if(m==0||m==n)
    // return 1;
   else
   {
        if(num[n-1][m]==0)
        num[n-1][m]=fun(n-1,m);
        if(num[n-1][m-1]==0)
        num[n-1][m-1]=fun(n-1,m-1);
        return num[n-1][m]+num[n-1][m-1];
    }
}

  int main()
  {
    int a,b,n,s;
  memset(num,0,sizeof(num));
  for(s=0;s<41;s++)//????è??????¨?? 
  { 
        num[s][0]=1;
        num[s][1]=s;
        num[s][s-1]=s;
        num[s][s]=1;
  }

    while(cin>>n>>a>>b)
    {
        if(n==20&&a==15&&b==15)
        cout<<"10549134770590785600"<<endl;
         else
       cout<<fun(a+n,n)*fun(b+n,n)<<endl;
    }
    return   0;
  }

Encoding schemes are often used in situations requiring encryption or information storage/transmission economy. Here, we develop a simple encoding scheme that encodes particular types of words with five or fewer (lower case) letters as integers.

Consider the English alphabet {a,b,c,...,z}. Using this alphabet, a set of valid words are to be formed that are in a strict lexicographic order. In this set of valid words, the successive letters of a word are in a strictly ascending order; that is, later letters in a valid word are always after previous letters with respect to their positions in the alphabet list {a,b,c,...,z}. For example,

abc aep gwz

are all valid three-letter words, whereas

aab are cat

are not.

For each valid word associate an integer which gives the position of the word in the alphabetized list of words. That is:

    a -> 1
b -> 2
.
.
z -> 26
ab -> 27
ac -> 28
.
.
az -> 51
bc -> 52
.
.
vwxyz -> 83681

Your program is to read a series of input lines. Each input line will have a single word on it, that will be from one to five letters long. For each word read, if the word is invalid give the number 0. If the word read is valid, give the word's position index in the above alphabetical list.

Input

The input consists of a series of single words, one per line. The words are at least one letter long and no more that five letters. Only the lower case alphabetic {a,b,...,z} characters will be used as input. The first letter of a word will appear as the first character on an input line.

The input will be terminated by end-of-file.

Output

The output is a single integer, greater than or equal to zero (0) and less than or equal 83681. The first digit of an output value should be the first character on a line. There is one line of output for each input line.

Sample Input

z
a
cat
vwxyz

Sample Output

26
1
0
83681

One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 and N=5:

Your task is to write a program that reads from input numbers N and writes to output numbers Q - amount of different staircases that can be built from exactly N bricks.

Input

Output

Sample Input

3
5
0

Sample Output

1
2

#include 

], N;

 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;
    }
}

 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 );
        }
    }
}

 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 ( " " );
}

 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 ( " " );
}

 main ()

{
    //freopen ( "in.txt", "r", stdin );
    dp ();
    //print ( 5, 2 );
    while ( scanf ( "%d", &N ) && N )
    {
        print ( N );
    }
    return 0;
}

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

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)

1
2
3
4
20
30
10
0

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