HDU 4135 Co-prime（容斥原理求互质数）

Co-prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1339 Accepted Submission(s): 495

Problem Description

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

21 10 23 15 5

Sample Output

Case #1: 5Case #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

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容斥原理说明http://www.cppblog.com/vici/archive/2011/09/05/155103.html

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <sstream>#include <vector>#include <queue>#include <set>#include <map>#include <ctime>using namespace std;typedef long long LL;LL solve (LL n, LL r) {        vector<LL> p;        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;        for (LL msk=1; msk<(1<<p.size()); ++msk) {               LL mult = 1,bits = 0;               for (LL i=0; i<(LL)p.size(); ++i)                       if (msk & (1<<i)) {                               ++bits;                               mult *= p[i];                       }                LL cur = r / mult;               if (bits % 2 == 1){                       sum += cur;                   }else{                       sum -= cur;                   }        }        return r - sum;}int main(){    LL a,b,n,t,g = 0;;scanf("%I64d",&t);while(t--){g++;scanf("%I64d%I64d%I64d",&a,&b,&n);printf("Case #%I64d: %I64d\n",g,solve(n,b)-solve(n,a-1));}return 0;}