HDU 1028 Ignatius and the Princess III (母函数的应用)

最新推荐文章于 2021-07-27 09:58:29 发布

原创最新推荐文章于 2021-07-27 09:58:29 发布 · 272 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#母函数

HDU 同时被 2 个专栏收录

18 篇文章

订阅专栏

母函数

1 篇文章

订阅专栏

本文介绍了一个整数划分问题，并通过一个具体的例子说明了如何计算给定正整数N的不同划分方式的数量。提供了完整的C++代码实现，展示了通过迭代过程找到所有可能的划分方法。

Ignatius and the Princess III(传送门)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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

ps: 整数划分的问题。

#include<iostream>
using namespace std;
const int _max = 10001;
int c1[_max], c2[_max];// c1是保存各项可以组合的数目，c2是中间量，保存每一次的情况
int main()
{
    int nNum;
    int i, j, k;
    while(cin>>nNum)
    {
        for(i = 0; i <= nNum; i++)
        {
            c1[i] = 1; //对于第一项多项式的各项因式
            c2[i] = 0;
        }
        for(i = 2; i <= nNum; i++)
        {//从第二项开始
            //模拟
            for(j = 0; j <= nNum; j++)//j 指向已经算出来的因式的指数，c2[j] 存放的是指数为 j 的项的系数
                for(k = 0; k + j <= nNum; k += i)
                {//k 代表下一个因式的指数，因式的指数是隔 i 递增的。
                // 另外，因式中，指数为 k 的项的系数为 1
                    c2[j+k] += c1[j];//对于 c1[j] * x^j * x^k = c1[j] * x^(j+k)
                                //就把 x^(j+k) 的系数 c1[j] 加到存放当前答案的数组 c2里面
                }
            for(j = 0; j <= nNum; ++j)
            {//把数组 c2 元素滚动到 数组 c1 去，以便递归进行乘法运算；另外，清空数组 c2
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        cout<<c1[nNum]<<endl;
    }
    return 0;
}