题目链接：HDU 1028

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

Solution

题意

给定 \(n\)，求 \(n\) 的划分数。

思路

普通母函数。母函数 \(G(x) = (1+x+x^2+...)(1+x^2+x^4+...)(1+x^3+x^6+...)...\)。

\((1+x+x^2+...)=(x^{0\times1}+x^{1\times1}+x^{2\times1}+...)\) 代表不用数字 \(1\)，用一次数字 \(1\)，用两次数字 \(1\)……

动态规划的版本见这里。

Code

#include <bits/stdc++.h>

using namespace std;

const int maxn = 200;

int c1[maxn], c2[maxn];

void init() {

    for(int i = 0; i < maxn; ++i) {

        c1[i] = 1;

        c2[i] = 0;

    }

    for(int i = 2; i < maxn; ++i) {

        for(int j = 0; j < maxn; ++j) {

            for(int k = 0; k + j < maxn; k += i) {

                c2[k + j] += c1[j];

            }

        }

        for(int j = 0; j < maxn; ++j) {

            c1[j] = c2[j];

            c2[j] = 0;

        }

    }

}

int main() {

    init();

    int n;

    while(~scanf("%d", &n)) {

        printf("%d\n", c1[n]);

    }

    return 0;

}