【整数划分+DP】HDU_1028_D

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22366    Accepted Submission(s): 15616

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

41020

 

Sample Output

542627

 

Author

Ignatius.L

 

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#include <bits/stdc++.h>using namespace std;int split(int n,int m){    if(n==1||m==1)        return 1;    if(m>n)        return split(n,n);    else if(n==m)        return split(n,m-1)+1;    else        return split(n,m-1)+split(n-m,m);}int dp[125][125];int main(){    int n;    while(~scanf("%d",&n)){//        int ans=split(n,n);//        printf("%d\n",ans);        for(int i=1;i<=n;i++){/*分解数i*/            for(int j=1;j<=n;j++){/*分解的数中最大的为j*/                if(i==1||j==1)                    dp[i][j]=1;                else if(j>i)                    dp[i][j]=dp[i][i];                else if(i==j)                    dp[i][j]=dp[i][j-1]+1;                else                    dp[i][j]=dp[i][j-1]+dp[i-j][j];            }        }        printf("%d\n",dp[n][n]);    }    return 0;}