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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 5010;
/**
*
* @param A
* @param B
* @return
*/
vector<int> add(vector<int> &A, vector<int> &B) {
if (A.size() < B.size()) return add(B, A);
int t = 0;
vector<int> C;
for (int i = 0; i < A.size(); i++) {
t += A[i];
if (i < B.size()) t += B[i];
C.push_back(t % 10);
t /= 10;
}
if (t) C.push_back(t);
return C;
}
vector<int> A, B, C;
int main() {
int n;
cin >> n;
if (n <= 2) {
printf("%d", n);
return 0;
}
//初始化一阶台阶步数
A.push_back(1);
//初始化二阶台阶步数
B.push_back(2);
//以后每阶台阶都是可以从i-1,i-2阶台阶迁移过来方法数就是两者的和
for (int i = 3; i <= n; i++) {
C = add(A, B);
//A<---B 方便下次计算
A.assign(B.begin(), B.end());
//B<---C 方便下次计算
B.assign(C.begin(), C.end());
}
//倒序输出结果
for (int i = B.size() - 1; i >= 0; i--)
printf("%d", B[i]);
return 0;
}