*HDU 1028 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): 19589    Accepted Submission(s): 13709

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

 

题意：

拆数共有多少总方案

代码：

/*//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;
}

#include<bits\stdc++.h>
using namespace std;
int c1[123],c2[123];
void solve()
{
    for(int i=0;i<=120;i++)
    {
        c1[i]=1;
        c2[i]=0;
    }
    for(int k=2;k<=120;k++)
    {
        for(int i=0;i<=120;i++)
        for(int j=0;j+i<=120;j+=k)
        c2[j+i]+=c1[i];
        for(int i=0;i<=120;i++)
        {
            c1[i]=c2[i];
            c2[i]=0;
        }
    }
}
int main()
{
    int n;
    solve();
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",c1[n]);
    }
    return 0;
}