HDU 1028（母函数）

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Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19181    Accepted Submission(s): 13477 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：给你一个正整数n，将该正整数拆分成若干整数的和，问你共有多少不同的拆分方法？

#include<bits/stdc++.h>

using namespace std;

int c1[],c2[];//c1用于记录各项前面的系数，c2用于记录中间值

int main()

{

    int n;

    while(~scanf("%d",&n))

    {

        //共有n个括号，先处理第一个括号，第一个括号里每一项的系数都是1；

        for(int i=;i<=n;i++)

        {

            c1[i]=;

            c2[i]=;

        }

        //因为共有n个括号，i表示正在处理第i个括号

        for(int i=;i<=n;i++)

        {

            //j表示正在处理第i个括号里的第就项，

            for(int j=;j<=(n);j++)

            {

                for(int k=;k+j<=n;k+=i)

                {

                    c2[j+k]+=c1[j];

                }

            }

            for(int j=;j<=(n);j++)

            {

                c1[j]=c2[j];

                c2[j]=;

            }

        }

        printf("%d\n",c1[n]);

    }

    return ;

}