Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6546    Accepted Submission(s): 4625


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
Author Ignatius.L 題目型別：普通型母函式 題目描述：問整數拆分有多少種形式 題目分析： 母函式又名生成函式，對於數列{ A(n) } n >= 0，∑(k>=0) A(k) * X^k = A(0)  + A(1) * X^1 + A(2) * X^2 +……+ A(k) * X ^ k 為數列{A(n)}的普通型母函式，簡稱普母函式。 記多重集 X = { n1x1, n2x2,……,nnxn}， x1,x2,x3,x4,……xn表示n個不同的元素。 ni(  1<= i <= n)為正整數或無窮，ni表示元素xi在計數問題中可取的最多的重數（無窮表示可取重數不受限制），稱  X = { n1x1, n2x2,……,nnxn} 為 x1,x2,x3,....,xn的多重集。 對於類似於多重集計數問題的組合問題，可以用普母函式來解決。 多重集中的每個元素，匯出一個相應的母函式。比如  nixi, ni = 2 xi = 4.那麼相應的母函式G(x) = 1 + x ^ 4 + x ^ 8 然後把生成的所有母函式相乘，即可求解問題。 模擬多項式相乘的時候，可以用 a[i] 的值 表示 x^i 的係數。 那麼 c * x^i  * d * x^j  就相當於 a[i] * b[j] ,結果為 r[i+j] = a[i] * b[j]。 程式碼如下：

#include <stdio.h>
#include <memory.h>
#define N 121

int result[N];
int temp[N];

void init(){
    int i,j,k;
    memset(result,0,sizeof(result));
    memset(temp,0,sizeof(temp));
    for( i = 0; i < N; i++) {
        result[i] = 1;
    }
    for( k = 2; k < N; k++){
        for( i = 0; i < N; i++){
            for( j = 0; j < N; j += k) {
                if( result[i] != 0 && i + j < N) {
                    temp[i+j] += result[i];
                }
            }
        }
        for( i = 0; i < N; i++){
            result[i] = temp[i];
            temp[i] = 0;
        }
    }
}


int main()
{
    init();
    int n;
    while(scanf("%d",&n) != EOF) {
        printf("%d\n",result[n]);
    }


    return 0;
}