HDU 1028

Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. 

"The second problem is, given an positive integer N, we define an equation like this: 
  N=a[1]+a[2]+a[3]+...+a[m]; 
  a[i]>0,1<=m<=N; 
My question is how many different equations you can find for a given N. 
For example, assume N is 4, we can find: 
  4 = 4; 
  4 = 3 + 1; 
  4 = 2 + 2; 
  4 = 2 + 1 + 1; 
  4 = 1 + 1 + 1 + 1; 
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!" 
 

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file. 
 

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found. 
 

Sample Input

4 10 20
 

Sample Output

5 42 627
 



经典题,递推递归。

首先,我们引进一个小小概念来方便描述吧,record[n][m]是把自然数划分成所有元素不大于m的分法,例如:

当n=4,m=1时,要求所有的元素都比m小,所以划分法只有1种:{1,1,1,1};

当n=4,m=2时,。。。。。。。。。。。。。。。。只有3种{1,1,1,1},{2,1,1},{2,2};

当n=4,m=3时,。。。。。。。。。。。。。。。。只有4种{1,1,1,1},{2,1,1},{2,2},{3,1};

当n=4,m=5时,。。。。。。。。。。。。。。。。只有5种{1,1,1,1},{2,1,1},{2,2},{3,1},{4};

从上面我们可以发现:当n==1||m==1时,只有一种分法;

当n<m时,由于分法不可能出现负数,所以record[n][m]=record[n][n];

当n==m时,那么就得分析是否要分出m这一个数,如果要分那就只有一种{m},要是不分,那就是把n分成不大于m-1的若干份;即record[n][n]=1+record[n][n-1];

当n>m时,那么就得分析是否要分出m这一个数,如果要分那就{{m},{x1,x2,x3..}}时n-m的分法record[n-m][m],要是不分,那就是把n分成不大于m-1的若干份;即record[n][n]=record[n-m][m]+record[n][m-1];

那么其递归式:

record[n][m]= 1 (n==1||m==1)

record[n][n] (n<m)

1+record[n][m-1] (n==m)

record[n-m][m]+record[n][m-1] (N>m)

;




#include<stdio.h>
#include<string.h>
int main()
{
    int i,j,n;
    int a[125][125];
    memset(a,0,sizeof(a));
    for(i=1;i<125;i++)
    {
        a[i][1]=a[1][i]=1;
    }
    for(i=2;i<125;i++)
    {
        for(j=2;j<125;j++)
        {
            if(i<j)
            {
                a[i][j]=a[i][i];
            }
            if(i==j)
            {
                a[i][j]=a[i][j-1]+1;
            }
            if(i>j)
            {
                a[i][j]=a[i-j][j]+a[i][j-1];
            }
        }
    }
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",a[n][n]);
    }

}


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