hdu4135 Co-prime【容斥原理】

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 626    Accepted Submission(s): 234 

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

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

 

Recommend

lcy

 

就是求区间与2互质的数的个数，就是简单容斥吧，用dfs做会比较好。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<cstdlib>
 #define ll long long
 using namespace std;

 ll ans,l,r,k;
 int num,cnt;
 ll a[],prime[];
 bool boo[];
 int Case;

 void init_prime()
 {
     for (int i=;i<=;i++)
     {
         if (!boo[i]) prime[++cnt]=i;
         for (int j=;j<=cnt&&prime[j]*i<=;j++)
         {
             boo[prime[j]*i]=;
             if (i%prime[j]==) break;
         }
     }
 }
 void devide_prime()
 {
     num=;
     for (int i=;i<=cnt;i++)
     {
         if (k%prime[i]==)
         {
             a[++num]=prime[i];
             while(k%prime[i]==) k/=prime[i];
         }
     }
     if (k>) a[++num]=k;
 }
 void query(int deep,ll shu,ll qz,ll zhi)
 {
     if (deep>num)
     {
         ans+=qz*zhi/shu;
         return;
     }
     query(deep+,shu*a[deep],-qz,zhi);
     query(deep+,shu,qz,zhi);
 }
 int main()
 {
     init_prime();
     int cas;
     scanf("%d",&cas);
     while(cas--)
     {
         scanf("%lld%lld%lld",&l,&r,&k);
         devide_prime();ans=;
         query(,,,r);
         query(,,-,l-);
         printf("Case #%d: %lld\n",++Case,ans);
     }
 }