問題描述

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

問題分析

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

6
5 + 1
4 + 2, 4 + 1 + 1
3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1
2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + 1

共11種。

主要是通過遞迴來實現，遞迴函式為int split(int n,int m) ，其中n是待劃分的正整數，m是劃分中最大的加數。返回值即為劃分的種數。
（1）當n=1或者m=1，split（n，m）=1;
（2）當m>n，最大加數不可能比要劃分的正整數大，所以，split（n，m）=split（n，n）
（3）當n=m，split（n，m）=split（n，m-1）+1。比如說：split(6,6)=1+split(6,5)。也就是說整數6的劃分可以分為6和最大加數為5的兩部分。
（4）當n>m，split（n，m）=split（n，m-1）+split（n-m，m）。比如說split（6，4）=split（6,3）+split（2,4）
其中：
split(6,3) :
3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1
2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + 1

split（2,4）：
4 + 2, 4 + 1 + 1
這些劃分方法有相同的整數m（這裡是4）。所以，把所有劃分都減去4，就相當於求正整數n-m的劃分種類數。

AC程式碼

# include<stdio.h>
# include<string.h>

#define MAX 125

int num[MAX][MAX];//使用陣列將已經計算過的值儲存起來，免得遞迴重複計算

int split(int n,int m);

int main()
{
    memset(num,0,sizeof(num));
    int N;
    while(scanf("%d",&N)!=EOF)
    {
        printf("%d\n",split(N,N));  
    }
    return 0;
}

int split(int n,int m)
{
    if(num[n][m])
        return num[n][m];
    else if(n==1||m==1)
        num[n][m]=1;
    else if(n>m)
        num[n][m]=split(n,m-1)+split(n-m,m);
    else if(n<m)
        num[n][m]=split(n,n);
    else if(n==m)
        num[n][m]=split(n,m-1)+1;
    return num[n][m];
}