2.大斐波那契数

#include<iostream>
#include<cstring>

using namespace std;

const int len = 1000;
//每行为一个斐波那契数
int arr[len][len]{ 0 };

int main() {
    arr[1][0] = 1;
    arr[2][0] = 1;

    int n, dight = 0, temp;
    //dight 记录位数
    cin >> n;
    for (size_t i = 3; i <= n; i++) {
        temp = 0;
        for (size_t j = 0; j <= dight; j++) {
            arr[i][j] += arr[i - 1][j] + arr[i - 2][j];
            temp = arr[i][j] / 10;
            arr[i][j] %= 10;

            if (temp) {
                arr[i][j + 1] = temp;
                if (j == dight) {
                    dight++;
                }
            }
            //cout << arr[i][j] << ' ';
        }
    }

    for (int j = dight; j >= 0; j--) {
        cout << arr[n][j];
    }
}

while True:
    try:
        n = int(input())
        temp =1;
        temp1 =1;
        if n <=2:
            res = 1;
        else :
            for i in range(2,n):
                res=temp+temp1
                temp = temp1;
                temp1 = res;
        print(res)
    except:
        break;