[Project Euler] Problem 27

Euler published the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40² + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

Using computers, the incredible formula  n² 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, 79 and 1601, is 126479.

Considering quadratics of the form:

n² + an + b, where |a| 1000 and |b| 1000

where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |4| = 4

Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.

只说一点，b一定是正素数

因为当n为0时，表达式的值为b

直接上代码

#include < iostream >
#include < cmath >
using namespace std;

int prime[ 168 ] = { 2 , 3 , 5 , 7 , 11 , 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 };

bool isPrime( int a);
int getLength( int i, int a[]);

int main(){
int a[ 168 ];
int length[ 168 ];
int max = 0 ;
int tmp;
for ( int i = 0 ; i < 168 ; i ++ ){
length[i] = getLength(i,a);
}
for ( int i = 0 ; i < 168 ; i ++ ){
if (length[i] > max){
max = length[i];
tmp = i;
}
}
cout << " a= " << a[tmp] << " \tb= " << prime[tmp] << " \tlength= " << length[tmp] << endl;
cout << " product= " << a[tmp] * prime[tmp] << endl;
return 0 ;
}

bool isPrime( int a){
bool tmp = true ;
int i = 0 ;
while (prime[i] <= sqrt(a)){
if (a % prime[i] == 0 ){
tmp = false ;
break ;
}
i ++ ;
}
return tmp;
}

int getLength( int i, int a[]){
int length = 0 ;
int at = 0 ;
for ( int j =- 999 ; j < 999 ; j ++ ){
int k = 0 ;
int tmp = 0 ;
while ((k * k + j * k + prime[i]) > 0 && isPrime(k * k + j * k + prime[i])){
tmp ++ ;
k ++ ;
}
if (tmp > length){
at = j;
length = tmp;
}
}
a[i] = at;
return length;
}

运行结果：