Ignatius and the Princess III(母函数或动态规划)

  HDOJ1028:

    

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 +
4000
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

 

//此体可以看成是分苹果的特殊情况，即把n个苹果放入n个盘中；
//不过需要开辟一个dp数组保存计算值，避免重复计算超时；
#include<stdio.h>
#include<string.h>
int dp[130][130];
int apple(int a,int b)
{
if(dp[a][b]!=-1)
return dp[a][b];
if(a==0||b==1)
dp[a][b]=1;
else
{
if(a<b)
dp[a][b]=apple(a,a);
else
dp[a][b]=apple(a,b-1)+apple(a-b,b);
}
return dp[a][b];
}

int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
memset(dp,-1,sizeof(dp));
printf("%d\n",apple(t,t));
}
return 0;
}

//母函数求解；
#include <iostream>
using namespace std;

int main()
{
int n；
int c1[130],c2[130];
//c1数组存储最终结果，c2存储中间量；
while(cin>>n)
{
for(int i=0;i<=n;i++)
{
c1[i]=1;
c2[i]=0;
}
for(int i=2;i<=n;i++)//i是第i个表达式；
{
for(int j=0;j<=n;j++)//j是第一个表达式中的取0,1...n;
for(int k=0;k+j<=n;k+=i)//k是第i个表达式中的取0,i,2*i,3*i...;
c2[k+j]+=c1[j];
for(int j=0;j<=n;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
cout<<c1
<<endl;
}
return 0;
}