題目鏈接：

D. Simple Subset

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

A tuple of positive integers?{x1,?x2,?...,?xk}?is called simple if for all pairs of positive integers?(i,??j)?(1??≤?i??<??j?≤?k),?xi??+??xj?is a prime.

You are given an array?a?with?n?positive integers?a1,??a2,??...,??an?(not necessary distinct). You want to find a simple subset of the array?awith the maximum size.

A prime number (or a prime) is a natural number greater than?1?that has no positive divisors other than?1?and itself.

Let's define a subset of the array?a?as a tuple that can be obtained from?a?by removing some (possibly all) elements of it.

?

Input

The first line contains integer?n?(1?≤?n?≤?1000) — the number of integers in the array?a.

The second line contains?n?integers?ai?(1?≤?ai?≤?106) — the elements of the array?a.

?

Output

On the first line print integer?m?— the maximum possible size of simple subset of?a.

On the second line print?m?integers?bl?— the elements of the simple subset of the array?a?with the maximum size.

If there is more than one solution you can print any of them. You can print the elements of the subset in any order.

?

Examples input 2
2 3 output 2
3 2 input 2
2 2 output 1
2 input 3
2 1 1 output 3
1 1 2 input 2
83 14 output 2
14 83


題意：

選一個最大的子序列,滿足這個序列里的任意兩個數的和是素數;

思路：

可以是一個最大完全數的題,也可以是水題,因為奇數+奇數=偶數,偶數+偶數=偶數,(1除外);所以最多有一個奇數和一個偶數;
我寫的分情況討論的代碼真是跟翔一樣;

AC代碼：

#include <bits/stdc++.h> using namespace std; const int N=2e6+6; typedef long long ll; int n,a[1010],flag[N]; int get_prime() {for(int i=2;i<N;i++){if(!flag[i]){for(int j=2*i;j<N;j+=i){flag[j]=1;}}} } queue<int>qu; void print() {printf("%d\n",qu.size());while(!qu.empty()){printf("%d ",qu.front());qu.pop();} }int main() {get_prime();int f=0;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]==1){f++;qu.push(1);}}if(f>1){for(int i=1;i<=n;i++){if(a[i]>1&&!flag[a[i]+1]){for(int j=i+1;j<=n;j++){if(a[j]>1&&!flag[a[i]+a[j]]&&!flag[a[j]+1]){qu.push(a[i]);qu.push(a[j]);print();return 0;}}}}for(int i=1;i<=n;i++){if(a[i]>1&&!flag[a[i]+1]){qu.push(a[i]);print();return 0;}}print();}else if(f==1){for(int i=1;i<=n;i++){if(a[i]>1&&!flag[a[i]+1]){for(int j=i+1;j<=n;j++){if(a[j]>1&&!flag[a[i]+a[j]]&&!flag[a[j]+1]){qu.push(a[i]);qu.push(a[j]);print();return 0;}}}}for(int i=1;i<=n;i++){if(a[i]>1){for(int j=i+1;j<=n;j++){if(a[j]>1&&!flag[a[i]+a[j]]){printf("2\n");printf("%d %d",a[i],a[j]);return 0;}}}}for(int i=1;i<=n;i++){if(a[i]>1&&!flag[a[i]+1]){qu.push(a[i]);print();return 0;}}print();}else{for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){if(!flag[a[i]+a[j]]){qu.push(a[i]);qu.push(a[j]);print();return 0;}}}printf("1\n");printf("%d",a[1]);}return 0; }

?

轉載于:https://www.cnblogs.com/zhangchengc919/p/5421033.html

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