項目歐拉＃27

Question

我挑戰自己在項目歐拉，但目前停留在問題27，其中的問題指出：

歐拉出版了顯着的二次公式：

N²+ N + 41但是，當n = 40時，402 + 40 + 41 = 40（40 + 1）+ 41可以被41整除，但是當n = 40時，402 + 40 + 41 = 40（40 + 1）當然當n = 41時，41 2 + 41 + 41明顯可被41整除。

使用計算機，發現令人難以置信的公式n²79n + 1601，它產生80個素數連續值n = 0到79.係數的乘積79和1601是126479.

考慮到quadratics形式：

n²+ an + b，其中| a | 1000和| b | 1000

其中| n |是例如n的模數/絕對值， | 11 | = 11和| 4 | = 4 查找的係數，a和b的產物，爲產生該二次表達式>開始，其中n = 0。

素數爲N，的連續值的最大數目我寫下面的代碼，這給了我很快的答案，但它是錯誤的（它吐出我（-951）*（-705）= 670455）。有人可以檢查我的代碼，看看我的錯誤在哪裏嗎？

#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; 
} 

Answer 1

isprime函數有個bug。

在你的函數中，你檢查所有的2310 * j + d [k]，其中j < int（sqrt（n）/ 2310））以確保目標n是素數。但是，還需要一個額外的條件，或者您將過度排除某些素數。例如，當a = 1，b = 41且n = 0時，函數將檢查41是否是從j = 0開始的素數。因此，41是否可以被2310 * 0 + d [7 ] = 41也被證實，這導致虛假的回報。

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; 
}