HDU 4135 Co-prime(容斥原理）

Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.
Input
The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10 15) and (1 <=N <= 10 9).
Output
For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.
Sample Input
2
1 10 2
3 15 5
Sample Output
Case #1: 5
Case #2: 10

Hint
In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

一道容斥的题，之前做过一道比这个难一些的，而其这个不需要去筛选一个范围内的数的素英制，所以求素因子的方法也不太一样，并且这个数的可以大到10e9次，就求素因子要注意一下

#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<vector>using namespace std;int factor[20];int k;bool prime[31623];int pri[3500];int l=0;void getPrime()//先筛素数{    memset(prime,true,sizeof(prime));    for(int i=2;i<=31622;i++)        if(prime[i])        {            pri[l++]=i;            for(int j=i+i;j<=31622;j+=i)                prime[j]=false;        }}void getFactor(int num)//求一个的数素因子{    k=0;    int m=num;    for(int i=0;i<l&&pri[i]*pri[i]<=num;i++)    {        if(m%pri[i]==0)        {            factor[k++]=pri[i];            while(m%pri[i]==0)                m/=pri[i];        }        if(m==1)break;    }    if(m>1)        factor[k++]=m;}long long get(long long limit,int state)//在一个状态组合下的在limit的范围内的个数{    int countt=0;    int v=1;    for(int i=0;i<k;i++)    {        if(state&(1<<i))        {            countt++;            v*=factor[i];        }    }    long long ans=limit/v;    if(countt%2)        return ans;    else        return -ans;}int main(){    getPrime();    int n;    long long l,r;    int num;    int t=1;    scanf("%d",&n);    while(n--)    {        scanf("%lld%lld%d",&l,&r,&num);        long long ans=r;        getFactor(num);        for(int s=1;s<(1<<k);s++)            ans-=get(r,s);        if(l>1)        {            long long ans2=l-1;            for(int s=1;s<(1<<k);s++)            ans2-=get(l-1,s);            ans-=ans2;        }        printf("Case #%d: %lld\n",t++,ans);    }    return 0;}