HDU1028

"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

感叹母函数的强大，第一次接触，明白其中的道理后还是很容易用代码实现的

转载一篇小飞大佬的文章
小飞_Xiaofei
http://blog.csdn.net/xiaofei_it/article/details/17042651

这里面介绍的非常详细，看完这个后就基本上懂得原理了，那就来看一下代码实现

对于这个题1~~n都可能用到

那我要用几个1，几个2，几个n都得计算进去，所以就得有n个多项式相乘

(1 + x^1 + x^2..+x^n)(1 + x^2 + x^4 + x^6 + ...)(1 + x^3 + x^6 +..)(..)(1 + x^n)
//第一个表达式(1 + x^1 + x^2..+x^n)中 x的指数代表【解中'1'的出现次数】 比如x^2 = x^(1 * 2) 这是'1'出现了两次 x^3 = x^(1 * 3) '1'出现3次

这样明白了多项式之后就是代码实现相乘的结果了

首先得准备好一个n1的数组存放当前各个系数（结果），n2存放当前结果（首端个多项式*下一个一个多项式后的结果）（首端个多项式，前两个多项式相乘都会留下一个多项式——首端多项式（自取名））

一个循环：循环剩下的每一个多项式

更新两个数组的之，为下一次相乘做准备

一个循环：循环循环首端多项式的每一个项

一个循环：循环相乘跳跃的步数

#include<iostream>#include<algorithm>#include<memory.h>using namespace std;int main(){int n1[125],n2[125];int n;while(cin>>n){//memset(n1,1,sizeof(n1));//用memset没法赋值为1？？？ //memset(n2,0,sizeof(n2));//cout<<n1[n]<<endl; for(int i = 0; i <= n; i++) //初始化 第一个表达式 目前所有指数项的系数都为1          {              n1[i] = 1;              n2[i] = 0;          }  for(int i = 2;i <= n;i++){for(int j = 0;j <= n;j++){for(int k = 0;j + k <= n;k += i){n2[j + k] += n1[j];}}for(int l = 0;l <= n;l++){n1[l] = n2[l];n2[l] = 0;}}cout<<n1[n]<<endl;}