Orac and Factors（递归）

Orac and Factors

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Orac is studying number theory, and he is interested in the properties of divisors.

For two positive integers a and b, a is a divisor of b if and only if there exists an integer c, such that a⋅c=b.

For n≥2, we will denote as f(n) the smallest positive divisor of n, except 1.

For example, f(7)=7,f(10)=2,f(35)=5.

For the fixed integer n, Orac decided to add f(n) to n.

For example, if he had an integer n=5, the new value of n will be equal to 10. And if he had an integer n=6, n will be changed to 8.

Orac loved it so much, so he decided to repeat this operation several times.
Now, for two positive integers n and k, Orac asked you to add f(n) to n exactly k times (note that n will change after each operation, so f(n) may change too) and tell him the final value of n.

For example, if Orac gives you n=5 and k=2, at first you should add f(5)=5 to n=5, so your new value of n will be equal to n=10, after that, you should add f(10)=2 to 10, so your new (and the final!) value of n will be equal to 12.
Orac may ask you these queries many times.

Input
The first line of the input is a single integer t (1≤t≤100): the number of times that Orac will ask you.
Each of the next t lines contains two positive integers n,k (2≤n≤106,1≤k≤109), corresponding to a query by Orac.
It is guaranteed that the total sum of n is at most 106.

Output
Print t lines, the i-th of them should contain the final value of n in the i-th query by Orac.

Example

inputCopy
3
5 1
8 2
3 4

outputCopy
10
12
12

Note
In the first query, n=5 and k=1. The divisors of 5 are 1 and 5, the smallest one except 1 is 5. Therefore, the only operation adds f(5)=5 to 5, and the result is 10.

In the second query, n=8 and k=2. The divisors of 8 are 1,2,4,8, where the smallest one except 1 is 2, then after one operation 8 turns into 8+(f(8)=2)=10. The divisors of 10 are 1,2,5,10, where the smallest one except 1 is 2, therefore the answer is 10+(f(10)=2)=12.

In the third query, n is changed as follows: 3→6→8→10→12.

基本思路

n每加一次f(n)后都需要重新计算一次新的n’和f(n’)【n’=n+f(n)】，所以递归公式为：return divisor(n’, k-1);
终止条件为:k==0;

源代码

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int divisor(int n, int k)
{
	if (k != 0)
	{
		for (int i = 2; i <= n; i++)
		{
			if (n % i == 0)
			{
				n = n + i;
				break;
			}
			else
				continue;
		}
		k--;
		return divisor(n, k);
	}
	else
	{
		return n;
	}
}
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		long int a, b;
		cin >> a >> b;
		cout << divisor(a, b) << endl;
	}
	return 0;
}