Problem 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

以下文章便於對整數劃分的演算法大體有一個瞭解

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整數劃分問題是將一個正整數n拆成一組數連加並等於n的形式，且這組數中的最大加數不大於n。
    如6的整數劃分為
    最大數  
    6         6
    5        5 + 1
    4         4 + 2, 4 + 1 + 1
    3         3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1

    2        2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 1
    1         1 + 1 + 1 + 1 + 1 + 1

    共11種。下面介紹一種通過遞迴方法得到一個正整數的劃分數。

    遞迴函式的宣告為 int split(int n, int m);其中n為要劃分的正整數，m是劃分中的最大加數(當m > n時，最大加數為n)，
    1 當n = 1或m = 1時，split的值為1，可根據上例看出，只有一個劃分1 或 1 + 1 + 1 + 1 + 1 + 1
    可用程式表示為if(n == 1 || m == 1) return 1;

    2 下面看一看m 和 n的關係。它們有三種關係
    (1) m > n
    在整數劃分中實際上最大加數不能大於n，因此在這種情況可以等價為split(n, n);
    可用程式表示為if(m > n) return split(n, n);    
    (2) m = n
    這種情況可用遞迴表示為split(n, m - 1) + 1，從以上例子中可以看出，就是最大加
    數為6和小於6的劃分之和
    用程式表示為if(m == n) return (split(n, m - 1) + 1);
    (3) m < n
    這是最一般的情況，在劃分的大多數時都是這種情況。
    從上例可以看出，設m = 4，那split(6, 4)的值是最大加數小於4劃分數和整數2的劃分數的和。
    因此，split(n, m)可表示為split(n, m - 1) + split(n - m, m)

    根據以上描述，可得源程式如下：


#include 

   int split(int n, int m)
    {
      if(n < 1 || m < 1) return 0;
      if(n == 1 || m == 1) return 1;
      if(n < m) return split(n, n);
      if(n == m) return (split(n, m - 1) + 1);
      if(n > m) return (split(n, m - 1) + split((n - m), m));
   }

int main()
{
      printf("12的劃分數: %d", split(12, 12));
    return 0;
}

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然而遞迴在本題會超時，不過上述的敘述加深了我對整數劃分的理解。

非遞迴的的程式碼：

#include <iostream>
using namespace std;

int main()
{
	freopen("D:\\input.txt","r",stdin);




	int n;
	int A[121][121];
	for(int i=0;i<121;i++)
	{
		for(int j=0;j<121;j++)
		{
			A[i][j]=0;
			A[i][1]=1;
		}
	}
	A[0][0]=1;
	A[1][0]=1;
    A[1][1]=1;
    /*A[2][1]=1;
    A[2][2]=2;
    A[3][1]=1;
    A[3][2]=2;
    A[3][3]=3;*/



	for(int i=2;i<121;i++)
	{
		for(int j=1;j<=i;j++)
		{
			if(i-j<=j)
			{
				A[i][j]=A[i][j-1]+A[i-j][i-j];
			}
			else
				A[i][j]=A[i][j-1]+A[i-j][j];
		}
	}
	while(cin>>n)
		cout<<A[n][n]<<endl;
}

Problem 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