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Accepted
Alex
Asked:2020-03-07 04:27:40 +0000 UTC2020-03-07 04:27:40 +0000 UTC

编写一个将数字排序成对的程序[关闭]

  • 772

有 1, 2, 3 ... N ( N <= 500000 ) 中的数字。编写一个程序，将数字分成对，使它们的总和是质数。我不知道如何正确执行此分配，请告诉我。据我所知，我写了，但竞选活动根本不是。

#include "stdafx.h"
#include <iostream>
using namespace std;

int main()
{
setlocale(LC_ALL, "rus");
int i,a;
int A[500000];
 for (i = 1; i <= 500000; i++)
 a = A[i];
 a += A[i + 1];
 if ((a%a == 1) && (a % 1 == 1))
     cout << a;
 system("pause");
}
c++

1 个回答

  1. Best Answer
    bool isprime(int thenumber)
{
    int isitprime = 1, loop;
    for (loop = 3; (isitprime) && (loop<(sqrt(thenumber) + 1)); loop += 2)
        isitprime = (thenumber % loop);
    return(isitprime);
}

#include <algorithm>
#include <vector>
void main(int argc, char**argv)
{
    int k = 0, n = argc > 1 ? atoi(argv[1]) : 500000;
    std::vector<int> x, primes = { 2 };
    for (int i = 3; i < n * 2; i += 2)
        if (isprime(i)) primes.push_back(i);
    for (int i = 0; i < n; i++) x.push_back(i + 1);
    for (auto i = x.begin(), end = x.end(); i < end; i++)
    {
        for (auto j = i + 1; j < end; j++)
            if (std::binary_search(primes.begin(), primes.end(), *i + *j))
                printf("%7d %7d\n", *i, *j), std::iter_swap(++i, j), j = end - 1, k++;
    }
    printf("%d\n", k * 2);
}

    超过10000还有多余的数字，好吧，也许应该是（ツ）

    • -1

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