• HDUOJ----Ignatius and the Princess III


    Ignatius and the Princess III

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 9893    Accepted Submission(s): 6996


    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
     
    Author
    Ignatius.L
     
    母函数.....对于任意一个数你   (1+x+x^2+x^3+x^4+x^5+x^6+x^7...+x^n)*(1+x^2+x^4+x^6+x^8+x^10+.....)*(1+x^3+.....);
     1 #include<iostream>
     2 #include<vector>
     3 using namespace std;
     4 int main()
     5 {
     6     int n,i,j,k;
     7     while(cin>>n)
     8     {
     9       vector<int>c1(n+1,1);
    10       vector<int>c2(n+1,0);
    11       for(i=2;i<=n;i++)
    12       {
    13          for(j=0;j<=n;j++)
    14          {
    15              for(k=0;k+j<=n;k+=i)
    16              {
    17                c2[j+k]+=c1[j];
    18              }
    19          }
    20          for(j=1;j<=n;j++)
    21          {
    22              c1[j]=c2[j];
    23              c2[j]=0;
    24          }
    25       }
    26       cout<<c1[n]<<endl;
    27     }
    28     return 0;
    29 }
    View Code
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  • 原文地址:https://www.cnblogs.com/gongxijun/p/3231020.html
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