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[][];

int main()

{

    int n,m;

        //dp[1][1]=1;

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

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

            {

                if(j==)

                    dp[i][j]=;

                else if(i==j)

                    dp[i][j]=dp[i][j-]+;

                else if(i>j)

                    dp[i][j]=dp[i][j-]+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 ;

}

巴特西

HDU_1028_Ignatius and the Princess III_（母函数，dp）

Ignatius and the Princess III

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