HDU 1028 Ignatius and the Princess III (生成函数/母函数)

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

1
2
3

4
10
20

Sample Output

1
2
3

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;
}