D - Ignatius and the Princess III （母函数）-优快云博客

本文链接：https://blog.youkuaiyun.com/stdio_xuege/article/details/95935866

博客围绕整数分解问题展开，给定正整数N，求其不同分解方程的数量。如N为4时有5种分解方式。输入为正整数N，输出为分解方程数量。还给出了用C语言实现的代码，通过多层循环计算结果。

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题目：

"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

题意：整数分解，问有多少种分解方法

分析；只用把硬币的面值写成从1-n的即可；

代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int num1[130],num2[130];
int n;
int main()
{
while(scanf("%d",&n)!=EOF)
{

memset(num1,0,sizeof(num1));
num1[0]=1;
for(int i=1;i<=n;i++)
{
for(int j=0;j<=n;j++)
{
for(int k=0;k<=n-j;k+=i)
{
num2[k+j]+=num1[j];
}
}
for(int j=0;j<=n;j++)
{
num1[j]=num2[j];
num2[j]=0;
}
}
int ans=0;
printf("%d\n",num1[n]);
}
return 0;

}