HDU - 4135 Co-prime (inclusion and exclusion)

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}.

---

Question is intended: to give you an interval [a, b], and a number n, to find the number of the n number of prime interval.
Outline of Solution: Find [1, b] and the number n minus the prime interval [1, a prime number n and interval). Direct violence certainly find time out, the use of inclusion and exclusion n (A∪B∪C ...) = n ( A) + n (B) + n © ... -n (A∩B) -n (A∩C) -n ( B∩C) ... + n (A∩B∩C) ... and the number n to identify prime interval, and then subtracting the number of values on the left and n is not a prime. Specifically look at the code, there are comments.

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
typedef long long ll;
ll solve(ll r,ll n){
    vector<ll> p;
    //找出n的所有素因子
    for(ll i=2;i*i<=n;++i){
        if(n%i==0){
            p.push_back(i);
            while(n%i==0){
                n/=i;
            }
        }
    }
    if(n>1)
        p.push_back(n);
    ll sum=0;
    ll s=p.size();
    //二进制枚举加容斥原理找出与n不互素的数的个数
    for(ll i=1;i<(1<<s);i++){
        ll q=1;
        ll cnt=0;
        for(ll j=0;j<s;j++){
            if((1<<j)&i){
                q*=p[j];
                cnt++;
            }
        }
        if(cnt&1)
            sum+=r/q;
        else
            sum-=r/q;
    }
    return r-sum;//用r减去互素的个数就是不互素的个数
}
int main()
{
    int t;
    ll a,b,n;
    scanf("%d",&t);
    int c=1;
    while(t--){
        scanf("%lld%lld%lld",&a,&b,&n);
        //a~b中与n互质的个数等于（1~b）中与n互质的数减去（1~a）中与n互质的数
        ll ans=solve(b,n)-solve(a-1,n);
        printf("Case #%d: %lld\n",c++,ans);
    }
}