C++ 最大子集，其中每對元素的和為質數-C++-PHP中文網

C++ 最大子集，其中每對元素的和為質數

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發布： 2023-08-26 08:05:17

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C++ 最大子集，其中每对元素的和为质数

從給定數組中找到最大的子集，其中每對元素的和是一個質數。假設最大元素是100000，例如−

Input: nums[ ] = { 3, 2, 1,1 }
Output: size = 3, subset = { 2, 1, 1 }
Explanation:
Subsets can be formed: {3, 2}, {2, 1} and { 2, 1, 1},
In {2, 1, 1} sum of pair (2,1) is 3 which is prime,
and the sum of pairs (1,1) is 2 which is also a prime number.

Input: nums[ ] = {1, 4, 3, 2}
Output: size = 2, subset = {1, 4}
Explanation:
subset can be formed: {1, 4}, {4, 3}, and {3, 2}
All are of size 2 so we can take any subset like 1 + 4 =5 which is a prime number.

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尋找解的方法

首先，要確定這對數是否為質數，我們需要檢查它們的和是奇數還是偶數，因為除了2以外，偶數都不是質數。而且，如果兩個數都是奇數或偶數，它們的和就可能是偶數。

在這個問題中，我們將取三個數，x、y和z，其中任兩個數應該是奇數或偶數。然後，我們將檢查這個子集是否包含質數和的數對，這可能是可能的，如果：

子集中包含一些1的數字和一些其他數字，其中NUM 1應該是質數。
或如果子集只包含兩個數，它們的和是質數。

範例

 #include <bits/stdc++.h>
using namespace std;
#define M 100001
bool check_prime[M] = { 0 };
int sieve_of_eratosthenes(){
    for (int p = 2; p * p < M; p++){
       // If it is not marked then mark
        if (check_prime[p] == 0){
            // Update all multiples of p
            for (int i = p * 2; i < M; i += p)
                check_prime[i] = 1;
        }
    }
    return 0;
}
int main(){
    sieve_of_eratosthenes();
    int nums[] = {  3, 2, 1, 1};
    int n = sizeof(nums) / sizeof(nums[0]);
        int ones = 0;
    for (int i = 0; i < n; i++)
        if (nums[i] == 1)
            ones++;
    // If ones are present and
    // elements greater than 0 are also present
    if (ones > 0){
        for (int i = 0; i < n; i++){
            //checking whether num + 1 is prime or not
            if ((nums[i] != 1) and (check_prime[nums[i] + 1] == 0)){
                cout << ones + 1 << endl;
                // printing all the ones present with nums[i]
                for (int j = 0; j < ones; j++)
                cout << 1 << " ";
                cout << nums[i] << endl;
                return 0;
            }
        }
    }
    // If subsets contains only 1&#39;s
    if (ones >= 2){
        cout << ones << endl;
        for (int i = 0; i < ones; i++)
            cout << 1 << " ";
        cout << endl;
        return 0;
    }
    // If no ones are present.
    for (int i = 0; i < n; i++){
        for (int j = i + 1; j < n; j++){
            // searching for pair of integer having sum prime.
            if (check_prime[nums[i] + nums[j]] == 0){
                cout << 2 << endl;
                cout << nums[i] << " " << nums[j] << endl;
                return 0;
            }
        }
    }
// If only one element is present in the array.
    cout << -1 << endl;
    return 0;
}

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