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43#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<string>
using namespace std;
typedef long long ll;
int main(){
int t;
cin >> t;
while(t--){
string n;
cin >> n;
int two = 0,three = 0;
ll sum = 0;
for(auto it : n){
if(it == '2') two++;
if(it == '3') three++;
sum += it - '0';
}
// at each operation either 2(earlier it was two it became 4 so net 2 added) will be added or
// 6(went from 3 to 9 so net 6) will be added
// so let the number after squaring digit has sum x so
// x - sum = some combination of 2 and 6
// 2*x1 + 6*x2 = y
// so if we use two nested loops to find such combination of x1 and x2
// then worst case time complexity will be n^2/4
bool flag = false;
for(int i = 0; i <= two; i++){
for(int j = 0; j <= three; j++){
ll newsum = sum + 2*i + 6*j;
if(newsum%9 == 0) flag = true;
}
if(flag) break;
}
if(flag) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}