这一题的数据很大，longlong都存不下... 记得换成unsign longlong

下面heky学长写的是高精度算法...

#include<stdio.h>
#include<string.h>
#include<stdlib.h> 
#include<math.h>

int main(){
	int n,i;scanf("%d",&n);
	unsigned long long sum=0;
	unsigned long long a=1;
	for(i=0;i<=n;i++){
		sum+=a;
		a*=2;
	}
	printf("%llu",sum);
	
    return 0;
}

#include<iostream>
#include<vector>
using namespace std;

vector<int> mul(vector<int> &A,int b) 
{
    vector<int> C;
    int t = 0;
    for (int i =0;i<A.size();i++)
    {
        t += A[i]*b;
        C.push_back(t%10);
        t /= 10;
    }
    while(t)
    {
        C.push_back(t%10);
        t /= 10;
    }
    return C;
}
vector<int> add(vector<int> &A,vector<int> &B) 
{
    vector<int> C;
    int t = 0;
    if (A.size() < B.size()) return add(B,A);
    for (int i = 0;i<A.size();i++)
    {
        t += A[i] + B[i];
        C.push_back(t%10);
        t /= 10;
    }
    while (t)
    {
        C.push_back(t%10);
        t /= 10;
    }
    return C;
}
int main() 
{
    int n;
    vector<int> B(1,1);
    cin >> n;
    for (int i = 1;i<=n;i++)
    {
        vector<int> A(1,1);
        for (int j =1;j<=i;j++)
        {
            A = mul(A,2);
        }
        B = add(B,A);
    }
    for (int i = B.size() - 1;i>= 0;i--) printf("%d",B[i]);
    return 0;
}

#include <stdio.h>
int main() {
    int n;
    scanf("%d", &n);
    unsigned long long sum = 0;
    unsigned long long term = 1;
    for (int i = 0; i <= n; i++) {
        sum += term;
        term <<= 1;  
    }
    printf("%llu\n", sum);
    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;

int main()
{   
    int n;
    cin >> n;
    unsigned long long sum = 0;
    for(int i = 0; i <= n; i++)
    {
        sum += powl(2,i);
    }
    cout << sum;
    return 0;
}

#include<stdio.h>

int main(){
	
	int n;
	scanf("%d",&n);
	
	unsigned long long sum=0;
	
	unsigned long long a=1;
	int i;
	for(i=0;i<=n;i++){
		sum+=a;
		a*=2;
	} 
	
	printf("%llu",sum);
	
	return 0;
	
}

#include <iostream>
using namespace std;
int main()
{
    unsigned long long ans = 1, last = 1;
    int n, i;
    cin >> n;
    for (i = 1; i <= n; i++)
    {
        ans += last * 2;
        last = last * 2;
    }

    cout << ans;
    return 0;
}