HDU 4135 Co-prime

Co-prime

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4135
64-bit integer IO format: %I64d      Java class name: Main

 

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 <= 1015) and (1 <=N <= 109).

 

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}.

Source

The Third Lebanese Collegiate Programming Contest

 

解题：容斥+数论

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
vector<LL>g;
LL solve(LL x,LL y) {
    g.clear();
    for(LL i = 2; i*i <= y; ++i) {
        if(y%i == 0) {
            while(y%i == 0) y /= i;
            for(int j = g.size()-1; j >= 0; --j)
                if(abs(i*g[j] <= x)) g.push_back(-i*g[j]);
            g.push_back(i);
        }
    }
    if(y > 1) {
        for(int i = g.size()-1; i >= 0; --i)
            if(abs(y*g[i]) <= x) g.push_back(-y*g[i]);
        g.push_back(y);
    }
    LL ret = 0;
    for(int i = g.size()-1; i >= 0; --i) ret += x/g[i];
    return x - ret;
}
int main() {
    int t,cs = 1;
    LL a,b,n;
    scanf("%d",&t);
    while(t--){
        scanf("%I64d%I64d%I64d",&a,&b,&n);
        printf("Case #%d: %I64d
",cs++,solve(b,n) - solve(a-1,n));
    }
    return 0;
}