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  • HDU 1028 Ignatius and the Princess III

    Ignatius and the Princess III

    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
    :简单的母函数问题
     
    #include<iostream>
    using namespace std;
    #define M 125
    int a[M],b[M];
    int main()
    {
        int n;
        while(scanf("%d",&n)!=EOF)
        {
            int i,j,k;
            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;
    }
    ^^^转载请注明出处~~~
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  • 原文地址:https://www.cnblogs.com/hsqdboke/p/2454220.html
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