ZJNU 2354 - Tribute

Considered separately k = 1 k = 2 and k> = 3 case

2 and 3 these two primes special encountered then a direct output line

For the "divine dissatisfaction about the number of m, the small m and a maximum ratio that" this description refers to itself and 1 m in addition to these two because the yard to find that the biggest

Can be found sqrt (m), 2 pairs always occurs according to the factor (except sqrt (m)), so finding a factor i will be the smallest m / i as the largest factor

If a prime number, an output

Type n and k

When k = 1, only once all tribute, go directly to the process described above

when k = 2, if the number is an even number, can "sum of two primes and even always split into" Output 2 (2 and 3 have special considerations) The

If it is an odd number, it is determined whether it is a prime number, is the output 1, or to find less than n largest prime number p, then n can be divided into p and np two portions (since p maximum, so np as small as possible, the maximum factors also can be as small as possible)

k = 3, if the number is an even number, can "sum of two primes and even always split into" Output 2 (ibid.) The

However, if it is odd, it is determined whether it is a prime number, it is the output 1, otherwise n is split into n-2 and 2 in two parts, since 2 is a prime number, n-2 in this case, an odd number, it is determined n-2 is not a prime number, 2 is output, otherwise, n can be divided into three portions, and two n-3, n-3 as an even number, according to the "even always split into a sum of two prime numbers" direct output 3

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int findm(int in){
 4     int i,d=sqrt(in);
 5     for(i=2;i<=d;i++)
 6         if(in%i==0)
 7             return in/i;
 8     return 1;
 9 }
10 bool isp(int in){
11     int i,d=sqrt(in);
12     for(i=3;i<=d;i+=2)
13         if(in%i==0)
14             return false;
15     return true;
16 }
17 int main(){
18     ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
19     int T,n,m,k;
20     cin>>T;
21     while(T--){
22         cin>>n>>k;
23         if(n==2||n==3)
24             cout<<1<<endl;
25         else{
26             if(k==1)
27                 cout<<findm(n)<<endl;
28             else if(k==2){
29                 if(n%2==1){
30                     if(isp(n))
31                         cout<<1<<endl;
32                     else{
33                         m=n-2 ;
 34 is                          the while (! ISP (m))
 35                              M- = 2 ; // n-odd, odd-numbered 2 are prime numbers outside, so every other -2 
36                          COUT << findm (nm) + . 1 << endl;
 37 [                      }
 38 is                  }
 39                  the else 
40                      COUT << 2 << endl;
 41 is              }
 42 is              the else  IF (K> = . 3 ) {
 43 is                  IF (n-% 2 == . 1 ) {
 44 is                      IF(isp(n))
45                         cout<<1<<endl;
46                     else{
47                         if(isp(n-2))
48                             cout<<2<<endl;
49                         else
50                             cout<<3<<endl;
51                     }
52                 }
53                 else
54                     cout<<2<<endl;
55             }
56         }
57     }
58     
59     return 0;
60 }