題目描述
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
input
3
5 1
8 2
3 4
output
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.
題目大意
給出數n和k次操作，每次操作的方法：n=n+n除了1以外的最小因子。
求最後的n爲多少。
題目分析
首先，我們可以發現：當n爲偶數時，n的最小因子2，而（n+2）的最小因子也爲2…所以，當n爲偶數時，答案即爲n+2*k。
而當n爲奇數時，它的最小因子也爲奇數，因此n+n的最小因子後爲偶數。於是也就轉化爲了偶數的情況。
代碼如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <unordered_set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=1e5+5;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,k;
cin>>n>>k;
if(n%2==0) //n爲偶數的情況
{
cout<<n+2*k<<endl;
continue;
}
int s;
for(int i=3;i<=n;i++) //n爲奇數的情況
if(n%i==0)
{
s=i;
break;
} //第一次操作之後，n也就變爲了偶數，所以可以再用偶數的情況處理剩下的k-1次操作
cout<<n+s+(k-1)*2<<endl;
}
return 0;
}