[ACM] POJ 1664 put apples (n the same ball into the same box of M)

Put the Apples

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25952		Accepted: 16509

Description

Put m the same apples on n the same plate, allow some plates to be empty, ask how many different ways? (denoted by k) 5,1,1 and 1,5,1 are the same kind of sub-method.

Input

The first line is the number of test data t (0 <= T <= 20). Each of the following lines contains two integers m and n, separated by a space. 1<=m,n<=10.

Output

For each set of data entered M and N, the corresponding k is output in one line.

Sample Input

17 3

Sample Output

8

Source

[email protected]

Put n the same ball into the same box of M, ask how many ways.

Analysis: Some topics can be translated into such topics, such as having n the same ball, m the same box, requires at least a K ball in each box, ask how many ways: http://blog.csdn.net/duanxian0621/article/ details/7864791 method is: In advance in each box has put the K ball, then also left n-k*m a small ball, and then converted in order to, put n-k*m the same ball into m the same box inside, how many ways.

Sub-case discussion: A[i][j] represents the number of ways I put a small ball into a J box

① when the ball is placed, the smallest number of the box is empty, then the equivalent, I put a small ball into the j-1 box, that is a[i][j-1].

② when the ball is placed, the box with the fewest number of balls is not empty, so that there is at least one ball in each box, then put a small ball in each box beforehand, and there is a[i-j] [j].

So, the recursive formula for a[i][j]= A[i][j-1]+a[i-j][j], here is i>=j.

When i<j, no matter how to put, there will be empty box, it is advisable to take away an empty box (guaranteed to have an empty box), and then put I the same ball into the remaining j-1 the same box, there are a[i][j-1] methods, so a[i][j]=a[i][j-1].

Total: n the same ball into the same box

When I>=j, A[i][j]=a[i][j-1]+a[i-j][j];

When i<, A[i][j]=a[i][j-1], (think about it, can also be written here A[i][j]=a[i][i], because a[i][i]=a[i][i+1]=a[i][i+2]=a[i][i+3] ...

Initialize the A array code as:

int A[MAXN][MAXN];

void Prepare ()

{

for (int i=0;i<m;i++)

A[i][1]=a[0][i]=1;

for (int i=1;i<maxn;i++)

for (int j=2;j<maxn;j++)

if (j<=i)

A[i][j]= (a[i-j][j]+a[i][j-1])%mod;

Else

A[i][j]=a[i][i]; In fact, it's a[i][j-1], but it's the same as A[i][i.

}

Code:

#include <iostream>using namespace Std;const int maxn=11;int a[maxn][maxn];void init () {for    (int i=1;i<=10 ; i++)        a[0][i]=1,a[i][1]=1;    for (int i=1;i<maxn;i++) for        (int j=2;j<=maxn;j++)            if (j<=i)                a[i][j]=a[i][j-1]+a[i-j][j];            else                a[i][j]=a[i][i];} int main () {    init ();    int t,n,m;    cin>>t;    while (t--)    {        cin>>n>>m;        cout<<a[n][m]<<endl;    }    return 0;}