hdu 1028(普通型母函数)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6546 Accepted Submission(s): 4625



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

题目类型：普通型母函数

题目描述：问整数拆分有多少种形式

题目分析：

母函数又名生成函数，对于数列{ A(n) } n >= 0，∑(k>=0) A(k) * X^k = A(0) + A(1) * X^1 + A(2) * X^2 +……+ A(k) * X ^ k
为数列{A(n)}的普通型母函数，简称普母函数。

记多重集 X = { n1x1, n2x2,……,nnxn}， x1,x2,x3,x4,……xn表示n个不同的元素。 ni( 1<= i <= n)为正整数或无穷，ni表示元素xi在计数问题中可取的最多的重数（无穷表示可取重数不受限制），称 X = { n1x1, n2x2,……,nnxn} 为 x1,x2,x3,....,xn的多重集。

对于类似于多重集计数问题的组合问题，可以用普母函数来解决。
多重集中的每个元素，导出一个相应的母函数。比如 nixi, ni = 2 xi = 4.那么相应的母函数G(x) = 1 + x ^ 4 + x ^ 8
然后把生成的所有母函数相乘，即可求解问题。

模拟多项式相乘的时候，可以用 a[i] 的值 表示 x^i 的系数。 那么 c * x^i * d * x^j 就相当于 a[i] * b[j] ,结果为 r[i+j] = a[i] * b[j]。

代码如下：

#include <stdio.h>
#include <memory.h>
#define N 121

int result
;
int temp
;

void init(){
int i,j,k;
memset(result,0,sizeof(result));
memset(temp,0,sizeof(temp));
for( i = 0; i < N; i++) {
result[i] = 1;
}
for( k = 2; k < N; k++){
for( i = 0; i < N; i++){
for( j = 0; j < N; j += k) {
if( result[i] != 0 && i + j < N) {
temp[i+j] += result[i];
}
}
}
for( i = 0; i < N; i++){
result[i] = temp[i];
temp[i] = 0;
}
}
}

int main()
{
init();
int n;
while(scanf("%d",&n) != EOF) {
printf("%d\n",result
);
}

return 0;
}