题解 | #吃糖果#

典型 斐波那契数列

动态规划 一步搞定

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100;
int f[N];
int n;

int main() {
    cin >> n;
    f[1] = 1;
    f[2] = 2;
    if (n == 1 || n == 2) cout << f[n];
    else {
        for (int i = 3; i <= n; i++)
            f[i] = f[i - 1] + f[i - 2];
        cout << f[n];
    }
}
// 64 位输出请用 printf("%lld")