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!"
看不懂英文可以直接看上面对数字4的拆解，简单的说就是对任意的数N，计算出有多少中拆解组合，注意4=3+1和4=1+3是一个组合。
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.
输入N。
Output
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.
输出拆解组合的数量P。
Sample Input
41020
Sample Output
542627
解题思路：
1、分解一下，采用背包的思想，求f(x)，比如f(4)，那么装背包时最大先用4来装，最大再用3，最大再用2...，最大用1。
2、这样可以建立一个方程，设 F(x, n)表示对于x，采用小于等于n的数字来拆解，有多少种解法。
3、那么就有 f(x) = F(x, x)，如f(6) = F(6, 6)，表示用小于等于6的数字来拆解的解法数。
4、先看一个具体的例子，F(6, 1) 表示6全部用1拆解，F(6, 2)表示用1、2拆解，那么F(6, 2) = F(6, 1) + F(4, 2)，所以状态转移方程：
F(x, n) = F(x, n-1) + F(x - n, n)
分析F(6, 2) = F(6, 1) + F(4, 2)：
这里6先用1来拆，那么只剩下用2来拆的情况，用2拆的时候，因为已经固定了拆掉1个2，那么还剩下4要用小于等于2的数来拆。
处理的时候需要先后处理F(1,1), F(2,1)...F(6,1), F(2,2),...F(6,2),...,F(6,6)。
代码如下：
#include<bits/stdc++.h>
#include<string>
using namespace std;
int main()
{
int a[200];
int n;
while(cin>>n)
{
memset(a,0,sizeof(a));
a[0]=1;
for(int i=1;i<=n;i++)
for(int j=i;j<=n;j++)
{
a[j]+=a[j-i];
}
cout<<a[n];
}
return 0;
}