TASK DEFINITION: This task is to find the number of 'For' loops present in the given cpp program.
PROBLEM: int sum=0,used[32770]={0};//????sum???????????used?i???i????????
void f(int n)//??f??n????????
{
	for(int i=2;i<=n;i++)
	{
		if(n%i==0&&used[i]==0)
		{
			if(i!=2)
			{
				for(int j=2;j<=i-1;j++)
					used[j]=1;
			}
			if(n/i==1)//??????????
			{
				sum++;
				break;
			}
			else
			{
				f(n/i);
			}
			memset(used,0,sizeof(used));//?used???????
		}
	}
}
int main()
{
	int m,a[100];
	cin>>m;
	for(int i=1;i<=m;i++)
	{
		cin>>a[i];
		if(a[i]==1)
			cout<<1<<endl;
		else
		{
			f(a[i]);
			cout<<sum<<endl;
			sum=0;//??
			memset(used,0,sizeof(used));
		}
	}
	return 0;
}

SOLUTION: 3

PROBLEM: void decprime(int begain,int a);
int sum=1;
int main()
{
	int n,a;
	int k;cin>>n;
	for(k=0;k<n;k++){
		cin>>a;
		 decprime(2,a) ;
		 cout<<sum<<endl;
		 sum=1;
	}
	return 0;
}
void decprime(int begain,int a)
{
	int i;
	 //int u=int(sqrt(double(a)));
//int m[10000]={};

	//if(a==2||a==3||a==5)return 1;
	//else{
		for(i=begain;i<=a;i++){
		if(a%i==0&&i<=a/i)
		{
			sum++;
			decprime(i,a/i);
		}
		if(i>a/i)break;
			//m[i]=decprime(a/i);
		//else m[i]=0;
//sum+=m[i];
		//}
	 //return sum+1;
	 }
		}


SOLUTION: 2

PROBLEM: int means(int,int);
int main()
{
	int n,i=0;
	cin>>n;
	while(i<n)
	{
		int a;
		cin>>a;
		cout<<means(a,a)<<endl;
		i++;
	}
	return 0;
}



int means(int a,int b)
{
	int i,sum=0;
	if(a!=1)
	{
		for(i=b;i>=2;i--)
		{
			if(a%i==0)
			{
				sum=sum+means(a/i,i);              //?i????a??????a/i???a/i??????i????????????
			}
		}
		return sum;                  //?????a???????????
	}
	else
	{
		return 1;                       //?a=1?????????1
	}
}

SOLUTION:
1