topic:
Idea:
According to the meaning of the question, we first find the number of corresponding n-level steps.
n = 1 f(n) = 1
n = 2 f(n) = 2
n = 3 f(n) = 3
n = 4 f(n) = 5
n = 5 f(n) = 8
...... ......
n = n f(n) = f(n - 2) + f(n - 1)
So through the listed examples, we can find the corresponding rules, and then derive the formula:
n = n f(n) = f(n - 2) + f(n - 1)
Therefore, we can make it clear that the meaning of i in dp[i] is the number of corresponding step plans, and record the corresponding i
The code is explained in detail below:
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <unordered_map>
#define endl '\n'
#define int long long
#define YES puts("YES")
#define NO puts("NO")
#define umap unordered_map
#define All(x) (x).begin(),(x).end()
#pragma GCC optimize(3,"Ofast","inline")
#define ___G std::ios::sync_with_stdio(false),cin.tie(0), cout.tie(0)
using namespace std;
const int N = 2e6 + 10,MOD = 10007;
int dp[N];
int f(int x)
{
// 如果相应台阶方案数计算过了
// 直接返回之前计算过的结果
if(dp[x]) return dp[x] % MOD;
return (dp[x] = f(x - 1) % MOD + f(x - 2) % MOD);
}
inline void solve()
{
// dp 初始化
dp[1] = 1;
dp[2] = 2;
int n;
cin >> n;
cout << f(n) % MOD << endl;
}
signed main()
{
// freopen("a.txt", "r", stdin);
___G;
int _t = 1;
// cin >> _t;
while (_t--)
{
solve();
}
return 0;
}