Problem

题目描述

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

41020

 

Sample Output

542627

题意不难理解 首先想到的就是递归了，很显然 4=4+0=3+1+0=2+2+0=2+1+1+0=1+1+1+1，其实就是4等于1加3，然后看3的分解，一步一步往下分解。

一开始以为很简单，可以建立一个二维数组a[n][i],n表示该输入的数，i表示该分组数里1的个数，其实这样还增大了递归的复杂性，最后还是要用递归一层一层的算。

实际代码

#include<stdio.h>#include<string.h>int dp[125][125];//dp[ n ][ m ]表示 n 表示成最大的数不超过 m 的方法数int calc( int n, int m ){    if( dp[n][m] != -1 )        return dp[ n ][ m ];    if(n < 1 || m < 1) return dp[ n ][ m ] = 0;    if(n == 1 || m == 1) return dp[ n ][ m ] = 1;    if(n < m) return dp[ n ][ m ] = calc( n, n );    if(n == m) return dp[ n ][ m ] = calc( n, m-1 ) + 1;    return dp[ n ][ m ] = calc( n, m-1 ) + calc( n-m, m );}int main( ){    int n;    memset( dp, -1, sizeof(dp) );            //将dp数组所有元素都初始化为-1    while( scanf("%d",&n) != EOF )      printf("%d\n",calc( n, n ) );    return 0;}

找了一下其他的解答，发现大部分用的母函数？？？

母函数的意思就是把n 用1,2,3,4,5`````来表示的种数，构造母函数：(1+x+x^2+x^3+`````)*(1+x^2+x^4+````)*(1+x^3+x^6+````)````

不是很理解额。。。