The task is to find the smallest number with given sum of digits as s and number of digits as d. print the smallest number if possible, else print -1.

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The task is to find the smallest number with given sum of digits as s and number of digits as d.
 print the smallest number if possible, else print -1.
Constraints:
1 ≤ T ≤ 100
1 ≤ s ≤ 100
1 ≤ d ≤ 6
This very simple approach with time complexity is 0.01.
#include <iostream>

#include <iostream>
using namespace std;
long int sumofdigit(long int n){
long int sum=0; 
while(n>0) { 
sum=sum+(n%10);
n/=10;    } 
//cout<<sum<<endl; 
return sum;}

int main() {
//code int t; cin>>t;
while(t--) {  
int s,d;  
cin>>s>>d;
long  int f=1,l=9,j=1;
if(s>d*9)  {    
cout<<-1<<endl;
j=0; }else
for(int i=0;i<d-1;++i)  {  
 f=f*10;    
l=(l*10)+9;
  }  

// cout<<f<<" "<<l<<endl;

for(long int i=f;i<=l;++i) {  
 long int sum=0;      
sum=sumofdigit(i);             
if(s==sum) {    
cout<<i<<endl;     
j=0;      
break; }

  }  
if(j)cout<<-1<<endl;
   }
return 0;
}