HDU 1028-Ignatius and the Princess III-母函数-整数拆分

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问题及代码：

Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1 Accepted Submission(s) : 1

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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

Sample Output

Author

Ignatius.L

/*    *Copyright (c)2015,烟台大学计算机与控制工程学院    *All rights reserved.    *文件名称：HDU.cpp    *作    者：单昕昕    *完成日期：2015年2月9日    *版 本 号：v1.0        */#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N=10010;//c1数组保存各项组合的数目c2数组是中间变量保存每一次的情况int c1[N],c2[N];int main(){    int n,i,j,k;    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<=n;i++)//因为是第一个表达式(1+x+x2+..xn)所以c1数组初始化为1，        {            c1[i]=1;            c2[i]=0;        }        for(i=2;i<=n;i++)//从2遍历，求每一个表达式        {            //j 从0到n遍历，这里j就是(前面i个表达式累乘的表达式)里第j个变量            for(j=0;j<=n;j++)            {                for(k=0;k+j<=n;k=k+i)//k表示的是第j个指数，因为第i个表达式的增量是i,所以k每次增i                {                    c2[k+j]=c2[k+j]+c1[j];                }            }            for(j=0;j<=n;j++)            {                c1[j]=c2[j];//因为c2每次是从一个表达式中开始的,把c2的值赋给c1,而把c2初始化为0.                c2[j]=0;            }        }        printf("%d\n",c1[n]);//X的n次方的系数即为所求    }    return 0;}

运行结果：

知识点总结：
母函数。

学习心得：

表示这题一开始以为是找规律的，然后试了几种都错了，后来一找度娘才发现要用母函数。

这母函数又是什么鬼呢，请见http://blog.csdn.net/mikasa3/article/details/43670817母函数详解。

组合数学的，长知识啊~~

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