UVA 10539 Almost Prime Numbers( 素数因子)

Almost prime numbers are the non-prime numbers which are divisible by only a single prime number. In this problem your job is to write a program which finds out the number of almost prime numbers within a certain range.

Input
First line of the input file contains an integer N (N <= 600) which indicates how many sets of inputs are there. Each of the next N lines make a single set of input. Each set contains two integer numbers low and high (0 < low <= high < 1012).

Output
For each line of input except the first line you should produce one line of output. This line contains a single integer, which indicates how many almost prime numbers are within the range (inclusive) low…high.

Sample InputSample Output

31 101 201 5

341

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;#define M 1000005typedef long long ll;int n,c;ll low,high;bool vis[M];ll prime[M/10];void init(){    int m=sqrt(1000000+0.5);    int i,j;    c=0;    for(i=2;i<=m;i++)    {        if(!vis[i])        {            for(j=i*i;j<=1000000;j+=i)            {                vis[j]=true;            }        }    }    for(i=2;i<=1000000;i++)    {        if(!vis[i])            prime[c++]=(ll)i;    }}int main(){    int i;    ll j;    init();    scanf("%d",&n);    while(n--)    {        scanf("%lld%lld",&low,&high);        ll ans=0;        for(i=0;i<c;i++)        {            if(prime[i]>high)                break;            for(j=prime[i]*prime[i];j<=high;j*=prime[i])            {                if(j>=low)                {                    ans++;                }            }        }        cout<<ans<<endl;    }    return 0;}