HDU 1028-Ignatius and the Princess III（拆分整数-母函数-无限次）

比眉伴天荒 2022-09-25 01:25 28阅读 0赞

# Ignatius and the Princess III #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 18862 Accepted Submission(s): 13247**

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

Recommend

We have carefully selected several similar problems for you: [1171][] [1085][] [1398][] [2152][] [1709][]

## 题目意思： ##

给定一个正整数N，将其分解成多个正整数相加的形式。

## 解题思路： ##

母函数，套个模板，可以看作是取数无限次，不像 [HDU 2082][] 是有限次取数。

#include <bits/stdc++.h>
    using namespace std;
    #define maxn 150
    
    void solve(int n)
    {
        int i,j,k;
        int a[maxn],b[maxn];
        memset(a,0,sizeof(a));//保存各项的系数
        memset(b,0,sizeof(b));//计算用中间数组
        a[0]=1;
        for(i=1; i<=n; ++i)
        {
            for(j=0; j<=n; ++j)//枚举多项式各项
                for(k=0; k*i+j<=n; ++k)
                    b[k*i+j]+=a[j];//每一项的系数累加
            for(j=0; j<=n; ++j)
                a[j]=b[j],b[j]=0;//更新各项系数，并清空中间数组方便下一轮计算
        }
        cout<<a[n]<<endl;
    }
    
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int t;
        while(cin>>t)
        {
            solve(t);
        }
        return 0;
    }
    
    /**
    4
    10
    20
    **/

[1171]: http://acm.hdu.edu.cn/showproblem.php?pid=1171
[1085]: http://acm.hdu.edu.cn/showproblem.php?pid=1085
[1398]: http://acm.hdu.edu.cn/showproblem.php?pid=1398
[2152]: http://acm.hdu.edu.cn/showproblem.php?pid=2152
[1709]: http://acm.hdu.edu.cn/showproblem.php?pid=1709
[HDU 2082]: http://blog.csdn.net/mikasa3/article/details/52121532