组合数学 - 母函数 + 模板题 ： 整数拆分问题

Ignatius and the Princess III#

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13129    Accepted Submission(s): 9285

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

 

 

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Mean: 

 给你一个n，问你n的拆分有多少种。

analyse:

 经典的母函数运用题，不解释。

Time complexity：O(n^3)

 

Source code：

 

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// Memory   Time // 1347K     0MS // by : Snarl_jsb // 2014-09-18-15.21 #include<algorithm> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<vector> #include<queue> #include<stack> #include<map> #include<string> #include<climits> #include<cmath> #define N 1000010 #define LL long long using namespace std; int c1[300],c2[300]; int main() {     ios_base::sync_with_stdio(false);     cin.tie(0); //    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin); //    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);     int n;     while(cin>>n)     {         memset(c1,0,sizeof(c1));         memset(c2,0,sizeof(c2));         for(int i=0;i<=n;++i)         {             c1[i]=1;         }         for(int i=2;i<=n;i++)         {             for(int j=0;j<=n;++j)             {                 for(int k=0;k<=n;k+=i)                 {                     c2[k+j]+=c1[j];                 }             }             for(int j=0;j<=n;++j)             {                 c1[j]=c2[j];                 c2[j]=0;             }         }         cout<<c1[n]<<endl;     }     return 0; }