方案一,按表格模拟

#include <bits/stdc++.h>
int main(){
    std::string s,ans; std::cin>>s;
    while(s.length()%3)s='0'+s;
    for(int i=0;i<s.length();i+=3){
        if(s[i]=='0'&&s[i+1]=='0'&&s[i+2]=='0')std::cout<<0;
        else if(s[i]=='0'&&s[i+1]=='0'&&s[i+2]=='1')std::cout<<1;
        else if(s[i]=='0'&&s[i+1]=='1'&&s[i+2]=='0')std::cout<<2;
        else if(s[i]=='0'&&s[i+1]=='1'&&s[i+2]=='1')std::cout<<3;
        else if(s[i]=='1'&&s[i+1]=='0'&&s[i+2]=='0')std::cout<<4;
        else if(s[i]=='1'&&s[i+1]=='0'&&s[i+2]=='1')std::cout<<5;
        else if(s[i]=='1'&&s[i+1]=='1'&&s[i+2]=='0')std::cout<<6;
        else if(s[i]=='1'&&s[i+1]=='1'&&s[i+2]=='1')std::cout<<7;}
    return 0;}

方案二,纯计算

#include <iostream>
#include <string>
int twoN(int n){
    int a=1;
    for(int i=0;i<n;i++)a*=2;
    return a;}
int main(){
    short a[105]={};std::string s; std::cin>>s;
    while(s.length()%3)s='0'+s;
    for(int i=0;i<s.length();i++)
        if(s[i]=='1')a[i/3]+=twoN(2-i%3);
    for(int i=0;i<s.length()/3;i++)std::cout<<a[i];
    return 0;}

#include<iostream>
using namespace std;
string s,ans;
int main()
{
    cin>>s;
    while(s.length()%3)s='0'+s;
    for(int i=0;i<s.length();i+=1)
    {
        if(i%3!=2)continue;
        if(s[i-2]=='0'&&s[i-1]=='0'&&s[i]=='0')cout<<0;
        if(s[i-2]=='0'&&s[i-1]=='0'&&s[i]=='1')cout<<1;
        if(s[i-2]=='0'&&s[i-1]=='1'&&s[i]=='0')cout<<2;
        if(s[i-2]=='0'&&s[i-1]=='1'&&s[i]=='1')cout<<3;
        if(s[i-2]=='1'&&s[i-1]=='0'&&s[i]=='0')cout<<4;
        if(s[i-2]=='1'&&s[i-1]=='0'&&s[i]=='1')cout<<5;
        if(s[i-2]=='1'&&s[i-1]=='1'&&s[i]=='0')cout<<6;
        if(s[i-2]=='1'&&s[i-1]=='1'&&s[i]=='1')cout<<7;
    }
    cout<<ans;
}

三位一截断，计算三位内的值

#include<bits/stdc++.h>
using namespace std;

int main(){
	string s,ans;
	cin>>s;
	int ls=s.size(),tmp=0;
	for(int i=ls-1;i>=0;i--){
		tmp=tmp+(s[i]-'0')*pow(2,(ls-1-i)%3);
		if((ls-1-i)%3==2 || i==0){
			ans=char(tmp+'0')+ans;
			tmp=0;
		}
	}
	cout<<ans;
	return 0;
}