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组合数学 - 母函数 + 模板题 ： 整数拆分问题

2024-07-23 04:47:05   218人阅读

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
 
 

 

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：

 

// 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 longusing 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;}

　　

组合数学 - 母函数 + 模板题 ： 整数拆分问题


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