Co-prime(容斥定理+数论)

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}. 

题意：

求区间[A,B]内和N互质的数的个数

分析：

可以先求区间[1,A-1],再求区间[1,B]，和N互质的个数，然后后者减前者即答案

每个区间可以用容斥做，将N分解质因数，判断区间范围内可以被任意素因数组合整除的个数，加加减减，最后总的个数减去记为互质的个数（不过多解释了）

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
int p[100],cnt;
void divide(int n){
    cnt = 0;
    for(int i = 2; i * i <= n; i++){
        if(n % i == 0){
            p[cnt++] = i;
            while(n % i == 0) n /= i;
        }
    }
    if(n > 1) p[cnt++] = n;
}
ll cal(ll s,int n){
    divide(n);
    ll ans = 0;
    for(int i = 1; i < (1 << cnt); i++){
        ll tmp = 1,num = 0;
        for(int j = 0; j < cnt; j++){
            if(i & (1 << j)){
                tmp *= p[j];
                num++;
            }
        }
        if(num & 1) ans += s / tmp;
        else ans -= s / tmp;
    }
    return s - ans;
}
int main(){
    int T,cas = 0;
    scanf("%d",&T);
    while(T--){
        ll A,B;
        int N;
        scanf("%lld%lld%d",&A,&B,&N);
        ll ans = 0;
        ans = cal(B,N) - cal(A-1,N);
        printf("Case #%d: %lld\n",++cas,ans);
    }
    return 0;
}