hdu1028Ignatius and the Princess III（母函数）

朱雀 2022-07-30 13:20 138阅读 0赞

# 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; }