Ignatius and the Princess III

Time Limit:1000MS  Memory Limit:65536K
Total Submit:86 Accepted:66

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
      就是这个题目上次我写整数划分的时候半天也没找到。。。直接用递归会超时！！
代买如下：
#include<cstdio>
int main()
{
    int i,j,n,c[121][121];
    for(i=1;i<121;i++)
        for(j=1;j<121;j++)
        {
            if(i==1||j==1)
                c[i][j]=1;
            else if(i<j)
                c[i][j]=c[i][i];
            else if(i==j)
                c[i][j]=c[i][j-1]+1;
            else
                c[i][j]=c[i][j-1]+c[i-j][j];
        }
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",c[n][n]);
    }
    return 0;
}