HDU 1028 Ignatius and the Princess III (动态规划)

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

最容易想到的就是直接递归，但是复杂度很高，可以用动态规划降低复杂度。

Code

#include <bits/stdc++.h>
using namespace std;
const int maxn = 150;

int dp[maxn][maxn];  // dp[i][j] 表示将i划分成最大数不超过j的划分数

void solve() {
    for(int i = 1; i < maxn; ++i) {
        for(int j = 1; j < maxn; ++j) {
            if(i == 1 || j == 1) {
                dp[i][j] = 1;
            } else if(i < j) {
                dp[i][j] = dp[i][i];
            } else if(i == j) {
                dp[i][j] = dp[i][j - 1] + 1;
            } else {
                // dp[i][j - 1]表示最大数不超过j-1的方案数, dp[i - j][j]表示拿出一个j后最大数不超过j的方案数
                dp[i][j] = dp[i][j - 1] + dp[i - j][j];
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    solve();
    int n;
    while(cin >> n) {
        cout << dp[n][n] << endl;
    }
    return 0;
}