HDU 1028 Ignatius and the Princess III

A - Ignatius and the Princess III

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status

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 
      
      
      
      

 
看上去貌似是考母函数之类的一道题，但是到现在并没有掌握母函数的知识，在这里占个位，以后补充。
这题的意思是问，拆分一个整数一共有多少种方法。
我的做法是 DP。
可得到下面三个状态转移方程。
dp[i][j]表示为用最大不超过j的整数去合并为i.

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

先DP打表，然后输出即可。

#include <stdio.h>
#define N 125
int n,dp[N][N];
int main()
{
        for(int i=1;i<=120;i++)
                        dp[i][1]=1;
        for(int j=1;j<=120;j++)
                        dp[1][j]=1;
                for(int i=1;i<=120;i++)
                        for(int j=1;j<=120;j++)
                                if(i<j)
                                        dp[i][j]=dp[i][i];
                                else if(i>j)
                                        dp[i][j]=dp[i-j][j]+dp[i][j-1];
                                else
                                        dp[i][j]=dp[i][j-1]+1;
        while(scanf("%d",&n)>0)
                printf("%d\n",dp[n][n]);
        return 0;
}

在DISCUSS里看到还有大牛用完全背包做的。

#include<cstdio>
#define N 125
int n,dp[N];
int main()
{
        dp [0] = 1;
        for(int i = 1;i<=120;i++)
                for(int j = i;j<=120;j++)
                        dp[j]+=dp[j-i];
        while(scanf("%d",&n)>0)
                printf("%d\n",dp[n]);
}