这是一开始我的代码，会超时，但是可以用来找出规律。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int f(ll n){
	if(n<=2) return 0;
	if(n%3==0){
		return f((n/3-1)/2+n/3)+1;
/*最小的数最小为1，剩余的可以取出给另外两个数，这样可以使得中间数最大。*/
	}
	else{//余1或2的情况
		return f((n/3-1)/2+1+n/3)+1;
	}
}
int main()
{
	int x;
	ll maxn=2,temp;
	cin>>x;
	for(ll i=3;;i++){
//		cout<<i<<" "<<p<<endl;
		temp=f(i);
		if(temp>x) {
			maxn=i-1;
			break;
		}
	}cout<<maxn;
	return 0;
}

通过对输入输出的分析，我发现f(x)=2*f(x-1)+2 ,函数返回为最终answer。故代码为:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll f(ll n){
	if(n==1) return 6;
	return 2*f(n-1)+2;
}
int main()
{
	int x;
	cin>>x;
	cout<<f(x);
	return 0;
}

#include
using namespace std;
int main()
{
    long long x;
    cin>>x;
    long long n=2;
    for(int i=0;i<x;i++)
    {
        n=2*n+2;
    }
    cout<<n;
    return 0;
}

AC代码，求赞

#include<iostream>
using namespace std;
int main()
{
    long long x; 
    cin>>x;
    long long n=2;
    for(int i=0;i<x;i++)
    { 
        n=2*n+2;
    } 
    cout<<n; 
    return 0;
}

#include <iostream> using namespace std; long long z = 2; long long f(int x) { z *= 2; z += 1 + 1; if (x == 1) { return z; } return f(x - 1);

} int main() { int s; cin >> s; cout << f(s); return 0; }

首先这道题很明显是通过次数求长度

题都读了吧）打个比方，切割次数为1时： 把长度2等分，这样左右都等，据题意，需要把左边这一段长分一个最短：“1”，那一直保留左面2等分长度减去1就是本次所得长

注：一直分到最后发现剩下的终止长度只能是2

规律：用最后的2加上1的和乘2，再用这个数加一乘二，直到X次用完就可得到绳长n

例：x=2时，直接[（2+1）*2 +1]*2

if(s==1)
{
    return (2+1)*2;
}
return (函数名(s-1)+1)*2;

别忘了函数定义用long long哦！

#include<iostream>
using namespace std;
#define POPOPOPOPOPOPOPOPOPOPOPOPOOPOPLOL return
#define QWERTYUIO_____________________________ for
#define ______________________________________________ int
#define IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIllllllllllllllllllllllllllllllllllIIIIIIIIIIIIIIIIIIIIIIIIII long long
______________________________________________ main()

{
    IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIllllllllllllllllllllllllllllllllllIIIIIIIIIIIIIIIIIIIIIIIIII x; 
    cin>>x;
    IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIllllllllllllllllllllllllllllllllllIIIIIIIIIIIIIIIIIIIIIIIIII n=2;
    QWERTYUIO_____________________________(______________________________________________ i=0;i<x;i++)
    { 
        n=2*n+2;
    } 
    cout<<n; 
    POPOPOPOPOPOPOPOPOPOPOPOPOOPOPLOL 0;
}`
ACtongguo

#include<iostream>
using namespace std;
#define POPOPOPOPOPOPOPOPOPOPOPOPOOPOPLOL return
#define QWERTYUIO_____________________________ for
#define ______________________________________________ int
#define IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIllllllllllllllllllllllllllllllllllIIIIIIIIIIIIIIIIIIIIIIIIII long long
______________________________________________ main()

{
    IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIllllllllllllllllllllllllllllllllllIIIIIIIIIIIIIIIIIIIIIIIIII x; 
    cin>>x;
    IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIllllllllllllllllllllllllllllllllllIIIIIIIIIIIIIIIIIIIIIIIIII n=2;
    QWERTYUIO_____________________________(______________________________________________ i=0;i<x;i++)
    { 
        n=2*n+2;
    } 
    cout<<n; 
    POPOPOPOPOPOPOPOPOPOPOPOPOOPOPLOL 0;
}

这妈呀

''' #include<iostream> using namespace std; int main() { long long x; cin>>x; long long n=2; for(int i=0;i<x;i++) { n=2*n+2; } cout<<n; return 0; } '''

ha,我用的是long int

虽是递归，但就如您所看到的 能省就省

#include <iostream>
using namespace std;long long x;
long long func(long long x){
if(x==1) return 6; return 2*func(x-1)+2;
}
int main(){
cin >> x; cout << func(x); return 0;
}