HDU 1028 Ignatius and the Princess III（dp 母函数）

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

 

代码是zxp那搬来的 完全不明觉厉= =

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
__int64 a[150];
int main()
{
    int i,j;
    a[0]=1;//a[1]=1;a[2]=2;a[3]=3;a[4]=5;a[5]=7;
    for(i=1;i<=120;i++)
    {
        for(j=0;i+j<=120;j++)
        {
            a[i+j]+=a[j];
        }
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
       printf("%I64d\n",a[n]);
    }
    return 0;
}

View Code

 

 

dp

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[200][200];
int fun(int n,int m)
{
    if(dp[n][m]!=-1) return dp[n][m];
    if(n<0||m<0) return 0;
    if(n==1||m==1) return  dp[n][m]=1;
    else if(m>n)   return  dp[n][n]=fun(n,n);
    else if(n==m)  return  dp[n][m]=fun(n,m-1)+1;
    else if(n>m)   return  dp[n][m]=fun(n-m,m)+fun(n,m-1);
}
int main()
{
    int n;
    int i,j;
    memset(dp,-1,sizeof(dp));
    while(scanf("%d",&n)!=EOF)
    {
        int ans=fun(n,n);
        printf("%d\n",ans);
    }
    return 0;
}

View Code

 

 

母函数

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int c1[200],c2[200];
void fun(int n)
{
    int i,j,k;
    memset(c1,0,sizeof(c1));
    memset(c2,0,sizeof(c2));
    for(i=0;i<=n;i++)
    {
        c1[i]=1;
    }
    for(i=2;i<=n;i++)
    {
        for(j=0;j<=n;j++)
        {
            for(k=0;k+j<=n;k+=i)
            {
               c2[j+k]+=c1[j];
            }
        }
        for(j=0;j<=n;j++)
        {
            c1[j]=c2[j];
            c2[j]=0;
        }
    }

}
int main()
{
    int n;
    int i,j,k;
    fun(120);
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",c1[n]);
    }
    return 0;
}