hdu 1028 Ignatius and the Princess III (n的划分)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26219    Accepted Submission(s): 18101

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

题目大意:

求n有几种划分(3=1+2和3=2+1是同一种划分方案)。

 

dp[i][j]表示i分为j块一共有几种方案。

那么，一般的，考虑划分的j块中有多少个1，接着用截边法处理:

若有0个1，把这j块都减一，转化为i-j分为j块，dp[i][j]+=dp[i-j][j];

若有1个1，把这j块都减一，转化为i-j分为j-1块，dp[i][j]+=dp[i-j][j-1];

一直考虑到有k个1即可。

每个数的划分数即为sum dp[i][]。

 

#include <cstdio>
#include <cstring>

using namespace std;

const int maxn=120;

//动规打表
int dp[maxn+5][maxn+5];

int sum[maxn+5];

int main()
{
    memset(dp,0,sizeof(dp));
    memset(sum,0,sizeof(sum));
    for(int i=1;i<=maxn;++i)
    {
        dp[i][1]=dp[i][i]=1;
        for(int j=2;j<=i-1;++j)
        {
            for(int k=0;k<=j;++k)
            {
                dp[i][j]+=dp[i-j][j-k];
            }
        }
        for(int j=1;j<=i;++j)
        {
            sum[i]+=dp[i][j];
        }
    }
    
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d
",sum[n]);
    }
    
    return 0;
}