I got a C++ problem.Let n be a positive integer and let S(n) denote the number of divisors of n. For example, S(1)- 1, S(4)-3, S(6)-4 A positive integer p is called antiprime if S(n)くS(p) for all positive n 〈P. In other words, an antiprime is a number that has a larger number of divisors than any number smaller than itself. Given a positive integer b, your program should output the largest antiprime that is less than or equal to b. Input specification: A positive integer b. For 30% of test cases, b <-100 For 80% of test cases, b < 1000 For 100% of test cases, b < 100000 Output specification: The largest antiprime less than or equal to b. Sample Input Sample Output 12 15 20 24 12 12 12 24

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Answer #1

Here i provide the code snapshot...

main.cpp #include iostream» 2 using namespace std; 3 bool prime(int n) int c-0; for(int 1-2;in/2;1++) 6 8 if((i%2)- 0) 10 12 13 14 15 16 17 return true; eise return falseHere is the output snapshots...

Here are the extra outputs...

Here is the code...

#include <iostream>
using namespace std;
bool prime(int n)
int c=0;
for(int i=2;i<n/2;i++)
{ c++;
if(c==0 && n!=1)
return true;
return false;
int main()
int input,k;
cin>>input; //Takes input from user....
int count=0,max=-1;
for(int i=1;i<=input;i++)
for(int j=1;j<=i;j++)
count++; //count the numbers of divisors
if(max < count) //takes the maximum number of divisors
{ if(!prime(i)) //and the number is antiprime
k=i; //store the number who have greatest number of divisors
//cout << "k value is"<<max;
cout << k<<endl;
//cout << max;
return 0;

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