Ignatius and the Princess III

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

41020

 

Sample Output

542627
母函数基础
#include<iostream>#include<stdio.h>using namespace std;int main(){    int N;    int c1[125],c2[125];    while(cin>>N)    {        int i,j,k;        for(i=0;i<=N;i++)//初始化第一个表达式的系数        {            c1[i]=1;            c2[i]=0;        }        for(i=2;i<=N;i++)        {//从第二个表达式开始，因为有无限制个，所以有n个表达式            for(j=0;j<=N;j++)            {//从累乘的表达式后的一个表达式第一个到最后一个                for(k=0;k+j<=N;k+=i)                {//k为第j个变量的指数，第i个表达式每次累加i                    c2[j+k]+=c1[j];                }            }            for(j=0;j<=N;j++)            {//滚动数组算完一个表达式后更新一次                c1[j]=c2[j];                c2[j]=0;            }        }        printf("%d\n",c1[N]);    }    return 0;}