题解 | 吃糖果

// 动态规划
// 也是斐波那契数列
#include <string.h>
#include <iostream>

using std::cout;
using std::cin;
using std::endl;

int dp[1024] = {0};

void init() {
    dp[0] = dp[1] = 1;
    dp[2] = 2;
    for (int i = 3; i < 1024; ++i)
        dp[i] = dp[i - 1] + dp[i - 2];
}

int main() {
    init();
    int N;
    cin >> N;
    cout << dp[N] << endl;

    return 0;
}