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：

1. 创建一个所求范围内质数真值数组
2. 偶数（2的倍数）全为false，奇数全为true
3. 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