Problem:
Score : 200 points

Problem Statement

Given are 1-digit positive integers a and b. Consider these two strings: the concatenation of b copies of the digit a, and the concatenation of a copies of the digit b. Which of these is lexicographically smaller?

Constraints

1 \leq a \leq 9

1 \leq b \leq 9

a and b are integers.

Input

Input is given from Standard Input in the following format:

a b

Output

Print the lexicographically smaller of the two strings. (If the two strings are equal, print one of them.)

Sample Input 1

4 3

Sample Output 1

3333

We have two strings 444 and 3333. Between them, 3333 is the lexicographically smaller.

Sample Input 2

7 7

Sample Output 2

7777777

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include<bits/stdc++.h>
using namespace std;
/**
********************************************************************
******************* Author:Bisnu sarkar ****************************
********************************************************************
**/
#define ull            unsigned long long
#define ll             long long
#define vi             vector<int>
#define pb             push_back
#define mp             make_pair
#define pii            pair<int,int>
#define vit            vector<int> :: iterator
#define sit            set<int> :: iterator
#define vrit           vector<int> :: reverse iterator
#define ff             first
#define ss             second
#define endl           '\n';
#define what_is(x)     cerr << #x << " is " << x << endl;
///*....Debugger....*/
#define error(args...) {vector <string> _v = split(#args, ','); err(_v.begin(), args); cout << endl;}
vector <string> split(const string &s, char c) {
    vector <string> v;
    stringstream ss(s); string x;
    while (getline(ss, x, c)) v.emplace_back(x);
    return move(v);
}
void err(vector <string> :: iterator it) {}
template <typename T, typename... Args>
void err(vector <string> :: iterator it, T a, Args... args) {
    cout << it -> substr((*it)[0] == ' ', it -> length()) << " = " << a << ", ";
    err(++it, args...);
}
 
int ini(){int x; scanf("%d",&x);return x;}
long long inl(){ long long x;scanf("%lld",&x); return x;}
double ind(){double x; scanf("%lf",&x);return x;}
#define Iin ini();
#define Lin inl();
#define Din ind();

int set_1(int n,int pos){return n = (n | (1<<pos));}
int reset_0(int n,int pos){return n= n & ~(1<<pos);}
bool check_bit(int n,int pos){return n = n & (1<<pos);}

int dx[8]= {1,0,-1,0,-1,-1,1,1};
int dy[8]= {0,1,0,-1,-1,1,-1,1};
const int N = (int) 1e6 + 5;
const int M = (int) 1e9 + 7;
const double pi=2* acos(0.0);
const double eps=1e-9;

int a[N];

int main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL);  
    int t=1,te=0;
   // cin>>t;
    while(t--){
        int x,y;

        cin>>x>>y;

        int mi=min(x,y);
        int ma=max(x,y);

        while(ma--){
            printf("%d",mi);
        }

        //printf("Case %d: ",++te);
    }
    return 0;
}