我在Project Euler中挑战自己，但目前仍然遇到问题27，其中问题表明：
欧拉发表了显着的二次公式：n²+ n + 41
事实证明，公式将为连续值n = 0到39产生40个素数。但是，当n = 40时，402 + 40 + 41 = 40（40 + 1）+ 41可被41整除，并且当n = 41时，41 + 41 + 41明显可被41整除。
使用计算机，发现令人难以置信的公式n²79n+ 1601，它为连续值n = 0到79产生80个素数。系数79和1601的乘积为126479。
考虑形式的二次方：
n²+ an + b，其中| a | 1000和| b | 1000
其中| n |是n的模量/绝对值 例如| 11 | = 11和| 4 | = 4 找到产生＆gt;的二次表达式的系数a和b的乘积。连续值n的最大素数，从n = 0开始。
我写了下面的代码，它给了我很快的答案，但它是错的（它吐了我（-951）*（-705）= 670455）。有人可以查看我的代码，看看我的错误在哪里？
#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;
}
答案 0 :(得分：0)
isprime功能有一个错误。
在你的函数中，你检查所有2310 * j + d [k]，其中j < int（sqrt（n）/ 2310））以确保目标n是素数。但是，附加条件是l <1。 sqrt（n）也是必需的，否则你将过度排除一些素数。
例如，当a = 1，b = 41且n = 0时，你的函数将检查41是否是从j = 0开始的素数。那么41是否可以被2310 * 0 + d [7]整除= 41也被验证，这导致错误的返回。
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;
}