小G的GCD

long long maxGCD(long long n) { long long ans = 0ll; for (long long i = 1; i <= n; i++) for (long long j = 1; j <= n; j++) ans = max(ans, GCD(i, j)); return ans; }

C++(clang++11) 解法, 执行用时: 2ms, 内存消耗: 380K, 提交时间: 2021-03-28 14:19:29

#include<bits/stdc++.h>
using namespace std;
int main()
{
    unsigned long long n;cin>>n;
    cout<<(int)(log(sqrt(5)*n+1.5)/log((1+sqrt(5))/2));
    return 0;
}

pypy3(pypy3.6.1) 解法, 执行用时: 45ms, 内存消耗: 18704K, 提交时间: 2021-03-27 17:20:03

if __name__=="__main__":
    n=int(input())
    a,b=1,1
    ans=2
    while(a+b<=n):
        a,b=a+b,a
        ans+=1
    print(ans)

Python3(3.9) 解法, 执行用时: 19ms, 内存消耗: 2808K, 提交时间: 2021-03-26 22:14:59

n = int(input())
a1 , a2 = 1 ,1
awn = 1
while a2<=n:
    a1 , a2 = a2 , a1 + a2
    awn += 1
print(awn)

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