*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[],c2[];
 void solve()
 {
     for(int i=;i<=;i++)
     {
         c1[i]=;
         c2[i]=;
     }
     for(int k=;k<=;k++)
     {
         for(int i=;i<=;i++)
         for(int j=;j+i<=;j+=k)
         c2[j+i]+=c1[i];
         for(int i=;i<=;i++)
         {
             c1[i]=c2[i];
             c2[i]=;
         }
     }
 }
 int main()
 {
     int n;
     solve();
     while(scanf("%d",&n)!=EOF)
     {
         printf("%d\n",c1[n]);
     }
     return ;
 }