求primes的算法

1、利用定义

质数：一个大于1的自然数，除了1和它本身外，不能被其他自然数整除。

int primeDef(int x){
int count = 0;
for(int key = 2; key <= x; key++){
int i = 2;
for(i = 2; i < key; i++)
if( key%i == 0 )  //若小于key的值能整除key,跳出循环,key不是质数
break;
if( i == key )  //若i完成整个循环,key是质数
count++;
}
return count;
}

static void test(void){
int n = primeDef(100);  //calcu the num of primes less than 100
cout<<n<<endl;
}

2、根据定理

定理：一个合数的最小质因数必小于其开平方根，反之则为质数。

int primeLemma(int x){
int count = x-1;  //initialize [2,x] as primes
for(int key = 2; key <= x; key++){
int min_factor = (int)sqrt(key);
int i = 2;
for(i = 2; i <= min_factor; i++)
if( key%i == 0 ){  //if can be divided, then count--
count--;
break;
}
}
return count;
}

3、优化算法

筛选法求质数：

创建一个所求范围内质数真值数组

偶数（2的倍数）全为false，奇数全为true

由最小质数开始筛选，其倍数必不为质数

int primeOpt(int x){
bool isPrime[x+1];

for(int i = 0; i < x-1; i+=2){
isPrime[i] = false;
isPrime[i+1] = true;
}

int n = sqrt(x);
int count = 0;

for(int key = 2; key <= n; key++){  //n为x可能的最小质因数
int i = 2*key;
for(i = 2*key; i <= x; i+= key){
if( isPrime[key] && isPrime[i] )  //若key为质数，则其小于x的倍数均为合数，因为有了key为其质因数
isPrime[i] = false;
}
}
for(int i = 0; i < x+1; i++){
if( isPrime[i] == true )
count++;
}
return count;
}

运行时间比较：

static void test(void){
clock_t t1,t2,t3;
int n = 100000;

t1 = clock();
int n1 = primeDef(n);
t1 = clock() - t1;

t2 = clock();
int n2 = primeLemma(n);
t2 = clock() - t2;

t3 = clock();
int n3 = primeOpt(n);
t3 = clock() - t3;

cout<<"nums:"<<n1<<" times: "<<t1<<" clicks"<<endl;
cout<<"nums:"<<n2<<" times: "<<t2<<" clicks"<<endl;
cout<<"nums:"<<n3<<" times: "<<t3<<" clicks"<<endl;
}

输出：

nums:9592 times: 1555 clicks

nums:9592 times: 11 clicks

nums:9592 times: 2 clicks

thk