hdu1028Ignatius and the Princess III（母函数）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17503    Accepted Submission(s): 12274

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

//整数拆分模板
#include <iostream>
using namespace std; const int lmax=10000;
//c1是用来存放展开式的系数的，而c2则是用来计算时保存的，
//他是用下标来控制每一项的位置，比如 c2[3] 就是 x^3 的系数。
//用c1保存，然后在计算时用c2来保存变化的值。
int c1[lmax+1],c2[lmax+1]; int main() { int n, i, j, k ;
           // 计算的方法还是模拟手动运算，一个括号一个括号的计算，
           // 从前往后
 while ( cin>>n ) {
                     //对于 1+x+x^2+x^3+ 他们所有的系数都是 1
                     // 而 c2全部被初始化为0是因为以后要用到 c2[i] += x ;
 for ( i=0; i<=n; i++ ) {
                                c1[i]=1;
                                c2[i]=0; }
                      //第一层循环是一共有 n 个小括号，而刚才已经算过一个了
                      //所以是从2 到 n
 for (i=2; i<=n; i++) {
                                 // 第二层循环是把每一个小括号里面的每一项，都要与前一个
                                 //小括号里面的每一项计算。
 for ( j=0; j<=n; j++ )
                                 //第三层小括号是要控制每一项里面 X 增加的比例
                                 // 这就是为什么要用 k+= i ;
 for ( k=0; k+j<=n; k+=i ) {
                                                 // 合并同类项，他们的系数要加在一起，所以是加法，呵呵。
                                                 // 刚开始看的时候就卡在这里了。
                                                 c2[ j+k] += c1[ j]; }
                               // 刷新一下数据，继续下一次计算，就是下一个括号里面的每一项。
 for ( j=0; j<=n; j++ ) {
                                          c1[j] = c2[j] ;
                                          c2[j] = 0 ; } }
                    cout<<c1[n]<<endl; } return 0; }