Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16394 Accepted Submission(s): 11552

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

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题目大意：

就是将一个数 n 拆分成无序可以重复的数，可以几种方法。。

解题思路：

母函数。。。

(1+x+x^2+x^3+.....+x^n)*(1+x^2+x^4+....)*(1+x^3+x^6+...)*...

直接上代码：（可以当作模板）

#include 
    
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             using namespace std;#define MM(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;const int maxn = 1e6+5;const int mod = 1e9+7;const double eps = 1e-8;const int INF = 0x3f3f3f3f;LL gcd(LL a, LL b){ if(b == 0) return a; return gcd(b, a%b);}int c1[maxn];///每一次存的多项式中的数int c2[maxn];///中间变量int n;///拆分的数/**首先对c1初始化，由第一个表达式(1+x+x^2+..x^n)初始化，从0到n的所有x前面的系数都初始化为1.**/void Init(){ for(int i=0; i<=n; i++) { c1[i] = 1; c2[i] = 0; }}int main(){ while(cin>>n) { Init(); for(int i=2; i<=n; i++)///控制一共循环几次，也就是每个多项式乘几次 { for(int j=0; j<=n; j++)///第一个表达式的系数 { for(int k=0; k+j<=n; k+=i)///我们只需要 n 前面的系数 { c2[k+j] += c1[j]; } } for(int j=0; j<=n; j++)///我们在用 c1[]当作第一个乘 { c1[j] = c2[j]; c2[j] = 0;///c2不需要 } } cout<
             
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