Hit the table to find the law!
Difficulties in the law of 2 m ==, violence hit the table is very important.
AC Code (15ms):
/*代码没注释，回看两行泪*/
#include <functional>
#include <algorithm>
#include <stdexcept>
#include <iostream>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <cctype>
#include <vector>
#include <string>
#include <bitset>
#include <cmath>
#include <queue>
#include <stdio.h>
#include <stack>
#include <ctime>
#include <list>
#include <map>
#include <set>
#include <assert.h>
#define fill(x,c) memset(x,c,sizeof(x))
using namespace std;
typedef long long ll;
int t;
string sum(int n,char c){
if(n<=0) return "";
string ans;
for(int i=0;i<n;i++) ans+=c;
return ans;
}
string solve(int m,int n){
if(m==1){
return sum(n,'a');
}else if(m==2){
string res;
vector<string> ans={"" ,
"a",
"ab",
"aab" ,
"aabb",
"aaaba" ,
"aaabab",
"aaababb",
"aaababbb",
"aaaababba",
"aaaababbaa",
"aaaababbaaa",
"aaaababbaaaa"};
if(n<13) return ans[n];
res = "aaaa";
string base = "babbaa";
int num = (n-4)/6;
int ys = (n-4)%6 ;
for(int i=0;i<num;i++){
res+=base;
}
res+=base.substr(0,ys);
return res;
}else{
int num_abc = n/3;
int ys = n%3;
string res;
string base = "abc";
for(int i=0;i<num_abc;i++){
res+=base;
}
res+=base.substr(0,ys);
return res;
}
}
int main(){
cin>>t;
int m,n,cnt=1;
while(t--){
cin>>m>>n;
string res = solve(m,n);
cout<<"Case #"<<cnt<<": "<<res<<endl;
cnt++;
}
return 0;
}