P1028

  一开始没看懂题,看了题解才明白的 = =。思路是,先找规律,会发现有重合部分,利用这些重合部分,写出递推公式。

  num = 0 时,只有 1 种组合;

  num = 1 时,只有 1 种组合;

  num = 2 时,有 12, 2 两种组合;

  num = 3 时,有 13, 3 两种组合(不能超过原数的一半);

  num = 4 时,有 124, 24, 14, 4 四种组合;

  ...

  如此下去,会发现 num 为奇数时等于 num - 1 的结果,而最基本的递推部分 num = 0, num = 1 是明确的。

  写出递推式:

$$

\left{

\begin{aligned}

f[i] &= 1, i = 1 ; \\

f[i] &= f[i/2] + f[i - 1] , i is even; \\

f[i] &= f[i - 1] ; i is odd.

\end{aligned}

\right.

$$

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i < b; i++)
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define index(a) (a - 'A')
#define transUpp(a) (a - 32)
#define transLow(a) (a + 32)
#define ll long long
#define PB push_back
int gcd(int a, int b){return b == 0 ? a : gcd(a%b, a);}
const int N = 100010;
int main()
{
    int f[N], num;
    cin >> num;
    f[0] = 1;
    for (int i = 1; i <= num; i++)
        f[i] = ((i + 1) & 1) * f[i >> 1] + f[i - 1];
    cout << f[num] << endl;
    return 0;
}

  

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转载自www.cnblogs.com/darkchii/p/9647882.html