Not able to figure out the code for the given problem and the logic

#1

#include
using namespace std;
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int query;
cin>>query;
int b[query];
int k[query];
for(int i=0;i<query;i++){
cin>>b[i]>>k[i];
}
for(int i=0;i<query;i++){
int flag=0;
int sum=0;
for(int j=0;j<n && sum<=b[i];j++){
if(b[i]==a[j]){
sum+=a[j];
}
if(i!=j){
flag++;
}
}

	if(k[i]==flag){
		cout<<"Yes"<<endl;
	}
	else cout<<"No"<<endl;

}
return 0;

}

#2

logic:
step1 : take value of n and take value of prices of each item in an array a[]
step 2: and then take the no of queries ‘q’
step 3: then make a loop for each query, and for each query take value of A and k ( A is units of money , k is no of atleast items from which she can choose from so that she can spend her whole money)
step 4: for each query check :

for(int j=0;j<n;j++){
            if((A%a[j])==0){   // if a[i] divides A completely then only she will be able to spend her whole money and she can buy this item.......
                k1++;
           }
        }

step 5 : check k1 and k

            if(k1>=k)  // k is atleast no of item that she can buy
			cout<<"Yes"<<endl;
	       	else
			cout<<"No"<<endl;

Reference Code:

i hope this help if you have further doubt, feel free to ask

#3

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