hdu 1028 Ignatius and the Princess III(分解正整数的方案数)

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于 2016-10-15 16:14:17 发布

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分类专栏：排列组合文章标签： ACM

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本文探讨了一个经典的数论问题——给定一个正整数N，求解所有可能的加法组合，使得这些组合的和等于N。通过一个高效的算法实现解答，并提供了完整的代码示例。

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Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19270 Accepted Submission(s): 13527

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

Sample Output

题意：分解n有多少种方案

思路：模板题

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
    int a[350],b[350],i,j,k,n;
    while(cin>>n&&n)
    {
        for(i=0;i<=n;i++){
            a[i]=1;
            b[i]=0;
        }
        for(i=2;i<=n;i++){
            for(j=0;j<=n;j++)
                for(k=0;k+j<=n;k+=i)
                    b[k+j]+=a[j];
            for(j=0;j<=n;j++){
                a[j]=b[j];
                b[j]=0;
            }
        }
        cout<<a[n]<<endl;
    }
    return 0;
}