HDU 1028（母函數），hdu1028函數

HDU 1028（母函數），hdu1028函數

Online Judge Online Exercise Online Teaching Online Contests Exercise Author F.A.Q
Hand In Hand
Online Acmers
Forum | Discuss
Statistical Charts Best Coder ^beta
VIP | STD Contests
Virtual Contests 
    DIY | Web-DIY ^beta
Recent Contests  jxust hsy
 Mail 0(0)
 Control Panel 
 Sign Out  

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

題意：給你一個正整數n，將該正整數拆分成若干整數的和，問你共有多少不同的拆分方法？

#include<bits/stdc++.h>
using namespace std;
int c1[205],c2[205];//c1用於記錄各項前面的系數，c2用於記錄中間值
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        //共有n個括號，先處理第一個括號，第一個括號裡每一項的系數都是1；
        for(int i=0;i<=n;i++)
        {
            c1[i]=1;
            c2[i]=0;
        }
        //因為共有n個括號，i表示正在處理第i個括號
        for(int i=2;i<=n;i++)
        {
            //j表示正在處理第i個括號裡的第就項，
            for(int j=0;j<=(n);j++)
            {
                for(int k=0;k+j<=n;k+=i)
                {
                    c2[j+k]+=c1[j];
                }
            }
            for(int j=0;j<=(n);j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        printf("%d\n",c1[n]);
    }
    return 0;
}