int main(int argc, char const *argv[])
{
int i;
int N = atoi(argv[1]);
int nodesAlloc = ((N-2)/16384) + 2;
head= ( seg * ) malloc(sizeof(seg));
pt=head;
for (i=1;i<nodesAlloc;i++) {
pt->next = ( seg *) malloc(sizeof (seg)); //allocate forward nodes
pt->prev = pt; //allocate backward nodes
pt=pt->next;
}
printf("Allocated %d nodes\n",i);
sieveOfE(N);
primecount = countPrimes(N)-1;
printf("Amount of primes = %d \n", primecount);
while(numbers != EOF){
printf("Enter your numbers or EOF:");
// if an odd integer is entered or eof is detected exit
if ( scanf("%d", &numbers) == EOF || numbers % 2 != 0 )
{
break;
}
else{goldbach(numbers);}
}
return 0;
}
int main(int argc, char *argv[])
{
int num, a, b, result;
while(scanf("%d", &num) != EOF)
{
switch(num)
{
case 0:
case 1:
case 2:
case 3:
case 4:
case 5:
case 6:
case 7:
printf("Impossible.\n");
break;
default:
result = goldbach(num % 2 ? num - 5 : num - 4, &a, &b);
if(result)
printf("2 %d %d %d\n", num % 2 ? 3 : 2, a, b);
}
}
return 0;
}
int main()
{
bool tag[MAXN];
memset(tag,false,sizeof(tag));
findPrimes(tag);
goldbach(tag);
return 0;
}
int main(){
int n, largest, smallest;
build();
while(scanf("%d", &n) == 1 && n){
printf("%d\n", goldbach(n));
}
return 0;
}
int main(){
// 1、输入一个数
printf("请输入一个偶数");
int num;
scanf("%d",&num);
// 2、判断是否为偶数
if(num%2!=0){
// 3.1、如果是奇数，则返回
return -1;
}
else{
// 3.2、如果是偶数，则继续
goldbach(num);
}
printf("\n");
return 0;
}
int
main()
{
const uint_fast64_t n = 10000;
uint_fast64_t i;
prime_init(n);
for (i = 3; i < n; i += 2)
{
if (prime(i))
continue;
if (!goldbach(i))
break;
}
printf("%lu\n", i);
return EXIT_SUCCESS;
}
int main()
{
goldbach(86);
return 0;
}
