Решение задачи 2^N с Acmp

    #include<vector>

#include<iostream>

#define PB push_back

#define ll long long

using namespace std;

typedef vector<int> lll;


istream& operator >>(istream& stream,lll& a){

	string s;stream>>s;

	for(int i=s.size()-1;i>=0;--i)a.PB(s[i]-'0');

	return stream;}

ostream& operator <<(ostream& stream,lll a){

	for(int i=a.size()-1;i>=0;--i)stream<<a[i];

	return stream;}

bool operator >(lll a,lll b){

	if(a.size()>b.size())return true;

	if(a.size()<b.size())return false;

	for(int i=a.size()-1;i>=0;--i){

		if(a[i]>b[i])return true;

		if(a[i]<b[i])return false;}

	return false;}

lll operator +(lll a,lll b){

	lll c;int i,r=0;

	for(i=0;i<a.size()||i<b.size();++i){

	if(i<a.size())r+=a[i];

	if(i<b.size())r+=b[i];

	c.PB(r%10);r/=10;}

	if(r)c.PB(r);

	return c;}

lll operator -(lll a,lll b){

	lll c;int i,r=0;

	for(i=0;i<a.size();++i){

	r+=a[i];

	if(i<b.size())r-=b[i];

	if(r<0){c.PB(r+10);r=-1;}

	else{c.PB(r%10);r=0;}}

	while(c.size()>1&&c.back()==0)c.pop_back();

	return c;}

lll operator *(lll a,ll b){

	lll c;int i;ll r=0;

	if(b==0){c.PB(0);return c;}

	for(i=0;i<a.size();++i){

	r+=a[i]*b;

	c.PB(r%10);r/=10;}

	while(r){c.PB(r%10);r/=10;}

	return c;}

ll operator % (lll a,ll b) {

	lll c; int i; ll r=0,t;

	for (i=a.size()-1;i>=0;--i) {

	r=r*10+a[i];t=r/b; r-=t*b;}

	while (c.size ()>1&&c[c.size()-1]==0) c.resize (c.size()-1);

	return r;}



int main() {

lll a;ll b,n; a.PB(1);b=2;

cin>>n;

for(int i=0;i<n;i++)a=a*b;

cout<<a;

return 0;}