计蒜客 斐波那契数列

#include 
#include 

using namespace std;

long f(long n)
{
    long f1 = 1;
    long f2 = 1;
    long f3 = 0;

    if (n == 1 || n == 2)
        return 1;

    for (long i = 2; i < n; i++)
    {
        f3 = (f1+f2)%1000000007;
        f1 = f2%1000000007;
        f2 = f3%1000000007;
    }
    return f3;
}

int main()
{
    long n;
    cin >> n;
    long result = f(n);
    cout << result << endl;
    return 0;
}

由于输出的 f mod 10e9+7，数过大，结果mod会溢出，需要中途mod：(a+b)%c = a%c + b%c