[2016-04-27] [Codeforces] [665d-simple subset]

• Time: 2016-04-27-15:16:14
• Title Number: [2016-04-27][codeforces][665d-simple subset]
• The main idea: Given the n number of set A, ask the subset of the maximum size, so that the sum of 22 of the sub-set is a prime, output and any one such subset
• Analysis:
  • At most one 1:
    • First take any 3 numbers from a
    • If there are at most 1 of these 3 numbers, 1
    • By the principle of tolerance, there must be two numbers of the sum is even, and the sum of the even greater than 2 (must not be a prime number), so the answer subset, more than 1 of the numbers can only have two,
    • Spread to the whole A, that is, if a is at most 1 1, the subset has a maximum of 2 digits, then the answer is
      1. The sum of no two numbers is a prime number,
      2. There is a sum of two numbers that are primes, (even if there are many pairs, only one pair can be output)
  • If more than 1, then the answer is
    1. It's all 1.
    2. In addition to 1, there is also a number, the number +1 is the prime

  
 

   
  

  
#include<cstdio>
   
  

  
#include<cstring>
   
  

  
using namespace std;
   
  

  
const int maxn = 1E3 + 10;
   
  

  
const int maxp = 1E6*2 + 10;
   
  

  
int isnpri[maxp],a[maxn];
   
  

  
void getpri(){
   
  

  
 memset   ( isnpri   0   sizeof   isnpri    
   
  

  
  for   ( int   i  =    2    ;   I  <   Maxp  +    10    ;    ++  i  ) {  
   
  

  
 if(isnpri[i]) continue;
   
  

  
 for(int j = i * 2 ; j < maxp; j += i){
   
  

  
 isnpri[j] = 1;
   
  

  
 }
   
  

  
 }
   
  

  
}
   
  

  
int main(){
   
  

  
 getpri();
   
  

  
 int n,cnt1 = 0;
   
  

  
 scanf("%d",&n);
   
  

  
 for(int i = 0 ; i < n ; ++i){
   
  

  
 scanf("%d",&a[i]); 
   
  

  
 if(a[i] == 1) ++cnt1;
   
  

  
 }
   
  

  
 if(cnt1 > 1){
   
  

  
  for   ( int   i  =    0    ;   I  <   n     ++  i  ) {  
   
  

  
  if   ( a  [ i      !=    1    &&    !  isnpri  [ a  [ i      +    1     
   
  

  
 printf("%d\n%d",cnt1 + 1,a[i]);
   
  

  
 for(int j = 0 ; j < cnt1;++j){
   
  

  
 printf(" 1");
   
  

  
 }
   
  

  
 return 0;
   
  

  
 }
   
  

  
 }
   
  

  
 printf("%d\n",cnt1);
   
  

  
 for(int j = 0 ; j < cnt1;++j){
   
  

  
 printf("1 ");
   
  

  
 }
   
  

  
 return 0;
   
  

  
 }
   
  

  

   
  

  
  for   ( int   i  =    0    ;   I  <   n     ++  i  ) {  
   
  

  
  for   ( int   J  =    0    ;   J  <   n     ++  j  ) {  
   
  

  
  if   ( i  !=   J  &&    !  isnpri  [ a  [ i      +   a  [ j  ]) { 
   
  

  
 printf("2\n%d %d",a[i] , a[j]);
   
  

  
 return 0;
   
  

  
 }
   
  

  
 }
   
  

  
 }
   
  

  
 printf("1\n%d",a[0]);
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-04-27] [Codeforces] [665d-simple subset]