for斐波那契数列

#include <iostream>

using namespace std;

int main(){
    int a= 1, b = 1;
    int n;
    
    cin >> n;
    
    for(int i = 0; i < n - 1; i ++){
        int c = a + b;
        a = b;
        b = c;
    }
    
    printf("%d", a);
    
    return 0;
}