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
4
10
20

Sample Output
5
42
627

題意:就是計算例如:誰加誰等於4
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
輸出5

原理:

第一項:1+x+x^2+x^3…………..
這每一項的意思：
0個1(x^0),1個1(x^1)，兩個1組成2(x^2),三個1組成3…….n個1組成n（x^n）
第二項：1+x^2+x^4…………..
0個2(x^0),1個2(x^2),兩個2(x^4)…………………….

程式碼:

# include <iostream>

using namespace std;

int main(){

    int n,i,j,k;
    int c1[130],c2[130];

    while(cin>>n){

        //對第一項進行賦值 
        for(i=0;i<=n;i++){
            c1[i] = 1;
            c2[i] = 0;
        }

        for(i=2;i<=n;i++){//代表第i項 

            for(j=0;j<=n;j++){//代表i-1項,其中,滿足項的係數j<=n的進行迴圈,這裡不一定是j-1的每一項 
                for(k=0;k+j<=n;k+=i){//i項符合條件的
                    c2[j+k]+= c1[j];//i-1項的每一個符合條件的與i項相乘 
                } 
            }

            for(k=0;k<=n;k++){
                c1[k] = c2[k];//把i與i-1項的乘積從臨時儲存到c2到儲存到c1中,等待i+1項與c1的成績 
                c2[k] = 0;
            }    
        } 
        cout<<c1[n]<<endl; 
    }
    return 0;
}