HDU 4135 Co-prime

 

Co-prime

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}. 
容斥
求[A B]中与n互质的个数  我们只需要求[A B]中和n不互质的个数 
比如sum[B]=[1 B]中和n不互质的个数
结果为B-sum[B]-(A-1-sum[A-1])
我们只需要求出n个质因数  根据容斥公式 A+B+C-A*B-A*C-B*C+A*B*C
用二进制生成子集来计算sum

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int N=1000000+10;
int prime[N];
ll a[N];
ll b[N];
ll t;
ll cnt;
void _prime(){
    memset(prime,0,sizeof(prime));
    for(int i=2;i*i<=N;i++){
        if(prime[i]==0){
            a[t++]=i;
            for(int j=i*i;j<=N;j=j+i)prime[j]=1;
        }
    }
}
void deal(ll n){
    for(int i=0;i<t&&a[i]*a[i]<=n;i++){
        if(n%a[i]==0){
            b[cnt++]=a[i];
            while(n%a[i]==0)n=n/a[i];
        }
    }
    if(n>1)b[cnt++]=n;
   // for(int i=0;i<cnt;i++)cout<<b[i]<<" ";
    //cout<<endl;
}
ll cal(ll n){
    ll ans=0;
    for(ll i=1;i<(ll)(1<<cnt);i++){
        ll flag=0;
        ll temp=1;
        for(int j=0;j<cnt;j++){
            if(i&(ll)(1<<j)){flag++;temp=temp*b[j];}
        }
        if(flag%2==1)ans=ans+n/temp;
        else
            ans=ans-n/temp;
    }
    return ans;
}
int main(){
    int tt;
    t=0;
    _prime();
    scanf("%d",&tt);
    ll A,B;
    ll n;
    int i=0;
    while(tt--){
        i++;
       scanf("%I64d%I64d%I64d",&A,&B,&n);
       cnt=0;
       deal(n);
       //cout<<cal(A-1)<<endl;
       //cout<<cal(B)<<endl;
       printf("Case #%d: %I64d\n",i,B-cal(B)-A+1+cal(A-1));
       //cout<<B-cal(B)-A+1+cal(A-1)<<endl;
    }
}