Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17917    Accepted Submission(s): 12558

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

看了题解，学了两种方法。一种是母函数，一种是dp。

母函数：组合数学方法，第一次接触。

　　此题构造的母函数(1+x^1+x^2+x^3...+x^n)(1+x^2+x^4+x^6...+x^2n).....

　　第一项表示(0个1，1个1，2个1，3个1...)，第二项表示(0个2，1个2，2个2，3个2，4个2...)以此类推。

　　展开后，每一项的指数表示划分的这个数，系数表示该数的划分数。

import java.util.*;
import java.io.*;

public class Main {

    public static int cal(int n)
    {
        int c1[]=new int [n+1];
        int c2[]=new int [n+1];
        for(int i=0;i<=n;i++)
        {
            c1[i]=1;
            c2[i]=0;
        }
        for(int i=2;i<=n;i++)
        {
            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;
            }
        }
        return c1[n];
    }
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n;
        while(in.hasNext())
        {
            n=in.nextInt();
            System.out.println(cal(n));
        }
    }

}

dp:

dp[i][j]表示i这个数划分为最大加数不超过j的划分数。

if(i>j)  dp[i][j]=dp[i][j-1]+dp[i-j][j];

else if(i==j)   dp[i][j]=dp[i][j-1]+1;

else if(i<j)   dp[i][j]=dp[i][i];

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;

int dp[150][150];

int main()
{
    int n,m;
        //dp[1][1]=1;
        for(int i=1; i<=150; i++)
            for(int j=1; j<=150; j++)
            {
                if(j==1)
                    dp[i][j]=1;
                else if(i==j)
                    dp[i][j]=dp[i][j-1]+1;
                else if(i>j)
                    dp[i][j]=dp[i][j-1]+dp[i-j][j];
                else if(i<j)
                    dp[i][j]=dp[i][i];
            }
        while(scanf("%d",&n)!=EOF)
        {
            printf("%d\n",dp[n][n]);
        }

    return 0;
}