hdu 1028Ignatius and the Princess III

http://acm.hdu.edu.cn/showproblem.php?pid=1028

and the Princess IIITime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7214    Accepted Submission(s): 5118

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

 模板题：

#include<stdio.h>
int main()
{
    int n,i,j,k;
    int sum[122];//sum[i]存储的是可以组合成i的所有总数 
    int temp[122];// 存储的是每一次的情况 
    while(~scanf("%d",&n))
    {
        
        
        for(i=0;i<=n;i++)
        {
            sum[i]=1;//初始化第一表达式，由数值1可以组合成0-n，这是一种组合方法。 
            temp[i]=0;
        }
        for(i=2;i<=n;i++)//i表示第i个表达式 
        {
            
            for(j=0;j<=n;j++)
            {
                for(k=0;j+k<=n;k+=i)
                {
                    temp[j+k]+=sum[j];
                    /*
                    这句话是最不好理解的
                    1.temp[j+k]每次加是叠加不同结果。
                    2.加上sum[j]其实是第一个表达式系数乘以第二表达式系数（此处为1） 
                    j+k是形成数的大小，也是函数里面的指数，指数相加，系数相乘。 
                    不懂，，请无视，同不懂 
                    
                    */ 
                }
            }
            for(j=0;j<=n;j++)
            {
                sum[j]=temp[j];
                temp[j]=0;
            }
        }
        printf("%d\n",sum[n]);
    }
}