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我在 Project Euler 中挑战自己，但目前卡在问题 27 上，其中问题指出：

欧拉发表了非凡的二次公式：

n² + n + 41

事实证明，该公式将为 n = 0 到 39 的连续值产生 40 个素数。但是，当 n = 40 时，402 + 40 + 41 = 40(40 + 1) + 41 可以被 41 整除，当然当 n = 41, 41² + 41 + 41 显然可以被 41 整除。

使用计算机，发现了令人难以置信的公式 n² 79n + 1601，它为 n = 0 到 79 的连续值产生了 80 个素数。系数 79 和 1601 的乘积是 126479。

考虑形式的二次方程：

n² + an + b，其中 |a| 1000 和 |b| 1000

其中 |n| 是 n 的模数/绝对值，例如 |11| = 11 和 |4| = 4 求二次表达式的系数 a 和 b 的乘积，从 n = 0 开始，该二次表达式产生 > 连续 n 值的最大素数数。

我编写了以下代码，它很快就给了我答案，但它是错误的（它让我吐了（-951）*（-705）= 670455）。有人可以检查我的代码以查看我的错误在哪里吗？

#include <iostream>
#include <vector>
#include <cmath>
#include <time.h>
using namespace std;

bool isprime(unsigned int n, int d[339]);
int main()
{
    clock_t t = clock();
    int c[] = {13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311};
    int result[4];
    result[3] = 0;
    for (int a = -999; a < 1000; a+=2)
    {
        for (int b = -999; b < 1000; b+=2)
        {
            bool prime;
            int n = 0, count = 0;
            do
            {
                prime = isprime(n*n + a*n + b, c);
                n++;
                count++;
            } while (prime);
            count--;
            n--;
            if (count > result[3])
            {
                result[0] = a;
                result[1] = b;
                result[2] = n;
                result[3] = count;
            }
        }
        if ((a+1) % 100 == 0)
            cout << a+1 << endl;
    }
    cout << result[0] << endl << result[1] << endl << result[2] << endl << result[3] << endl << clock()-t;
    cin >> result[0];
    return 0;
}

bool isprime(unsigned int n, int d[339])
{
    int j = 0, l;
    if ((n == 2) || (n == 3) || (n == 5) || (n == 7) || (n == 11))
        return 1;
    if ((n % 2 == 0) || (n % 3 == 0) || (n % 5 == 0) || (n % 7 == 0) || (n % 11 == 0))
        return 0;
    while (j <= int (sqrt(n) / 2310))
    {
        for (int k = 0; k < 339; k++)
        {
            l = 2310 * j + d[k];
            if (n % l == 0)
               return 0;
        }
        j++;
    }
    return 1;
}
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1 回答 1

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isprime 函数有一个错误。

在您的函数中，您检查所有 2310 * j + d[k] 其中 j < int (sqrt(n) / 2310)) 以确保目标 n 是质数。但是，还需要 l < sqrt(n) 的附加条件，否则您将过度排除某些素数。

例如，当 a = 1、b = 41 和 n = 0 时，您的函数将检查 41 是否是从 j = 0 开始的素数。所以 41 是否可以被 2310 * 0 + d[7] = 41 整除是也经过验证，导致虚假退货。

This version should be correct:
bool isprime(unsigned int n, int d[])
{
    int j = 0, l;
    if ((n == 2) || (n == 3) || (n == 5) || (n == 7) || (n == 11))
        return 1;
    if ((n % 2 == 0) || (n % 3 == 0) || (n % 5 == 0) || (n % 7 == 0) || (n % 11 == 0))
        return 0;
    double root = sqrt(n);
    while (j <= int (root / 2310))
    {
        for (int k = 0; k < 339; k++)
        {
            l = 2310 * j + d[k];
            if (l < root && n % l == 0)
                return 0;
        }
        j++;
    }
    return 1;
}
于 2013-04-06T01:17:29.660 回答