Problem

　　https://codeforces.com/problemset/problem/1620/C

　　

 

 

 　　

 

 

 　　

       

Solution

　　To solve the problem,we have to bear one thing in mind:a series of asterisks has to be considered as a whole.

　　So we come to the easier part.Starting from the end of the original string,we find that a series of t1 asterisks has t1*k+1 possibilities of "b"s and we only have to calculate how many "b"s a series of asterisks means.

　　It's quite like a number game.For example: **a***,k=3,x=20------》7|10 and x-1=19-------》19=10*1+9---------》babbbbbbbbb.

　　So Ac code is:

#include<bits/stdc++.h>
using namespace std;
#define N 200005
#define mp make_pair
#define ll long long
ll t;
ll n,k,x;
//void test(string s) {
//    getchar();
//    reverse(s.begin(),s.end());
//    cout<<s<<endl;
//}
int main() {
    cin>>t;
    while(t--) {
        string s;
        cin>>n>>k>>x;
        cin>>s;
        x--;
        reverse(s.begin(),s.end());
        ll point = 0;
        while(point<s.length()) {
            if(s[point]=='*') {
                ll num = 0;
                ll st = point;
                while(s[point]=='*') {
                    num++;
                    point++;
                }
                ll en = point;
                string s1 = "";
                ll mod  = num*k+1;
                for(ll i=1; i<=x%mod; i++) {
                    s1.append("b");
                }
                s = s.substr(0,st-0)+s1+s.substr(en,s.length()-en);
                point = st+s1.length();
                x/=mod;
            } else {
                point++;
            }
        }
        reverse(s.begin(),s.end());
        string s1="";
        ll p=0;
        while(p<s.length()) {
            if(s[p]=='*') p++;
            else s1+=s[p++];
        }

        cout<<s1<<endl;
    }
    return 0;
}