Hdu 1028 Ignatius and the Princess III

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
题意就是将给定整数n分成任意整数的和，求分法。
本代码使用了母函数思想。

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <complex>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <ctime>#include <cassert>using namespace std;const int MAXN = 10000;int main(){    int n;    while(~scanf("%d",&n)){        int c1[MAXN],c2[MAXN];///c1数组存的是当前得到多项式中每一项前的系数,c2是活动数组，        for(int i=0;i<=n;i++){///第一个循环是将第一个括号中多项式中每一项前系数归1，（1+x+x2+x3+x4+...xn）,            c1[i]=1;            c2[i]=0;        }        for(int i=2;i<=n;i++){///代表第二个括号开始到第n个括号            for(int j=0;j<=n;j++)///j代表之前已经得到的多项式的指数，c1[j]代表系数            for(int k=0;k+j<=n;k+=i){///代表当前括号中的指数，因为是第i个括号，所以每次递加i                c2[j+k]+=c1[j];      ///当前括号指数为k，之前多项式指数为j，根据x1*x2=x1+2=x3,所以得到的指数为j+k，系数需要继承c1[j]的系数，同时加上可能第二次得到的系数，比如x1*x3=x2*x2=x4,系数叠加。            }            for(int j=0;j<=n;j++){                c1[j]=c2[j];         ///当前系数在下一个括号前转移到c1数组中，代表已经得到的系数。                c2[j]=0;             ///活动数组归0。            }        }        cout<<c1[n]<<endl;    }    return 0;}

大家还可以看看这篇博客。
http://blog.csdn.net/vsooda/article/details/7975485