hdu4135 Co-prime容斥原理

Posted 2020-10-11 Kaiser

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。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #define ll long long
 8 using namespace std;
 9 
10 ll ans,l,r,k;
11 int num,cnt;
12 ll a[107],prime[100007];
13 bool boo[100007];
14 int Case;
15 
16 void init_prime()
17 {
18     for (int i=2;i<=100000;i++)
19     {
20         if (!boo[i]) prime[++cnt]=i;
21         for (int j=1;j<=cnt&&prime[j]*i<=100000;j++)
22         {
23             boo[prime[j]*i]=1;
24             if (i%prime[j]==0) break;
25         }
26     }
27 }
28 void devide_prime()
29 {
30     num=0;
31     for (int i=1;i<=cnt;i++)
32     {
33         if (k%prime[i]==0)
34         {
35             a[++num]=prime[i];
36             while(k%prime[i]==0) k/=prime[i];
37         }
38     }
39     if (k>1) a[++num]=k;
40 }
41 void query(int deep,ll shu,ll qz,ll zhi)
42 {
43     if (deep>num)
44     {
45         ans+=qz*zhi/shu;
46         return;
47     }
48     query(deep+1,shu*a[deep],-qz,zhi);
49     query(deep+1,shu,qz,zhi);
50 }
51 int main()
52 {
53     init_prime();
54     int cas;
55     scanf("%d",&cas);
56     while(cas--)
57     {
58         scanf("%lld%lld%lld",&l,&r,&k);
59         devide_prime();ans=0;
60         query(1,1,1,r);
61         query(1,1,-1,l-1);
62         printf("Case #%d: %lld\n",++Case,ans);
63     }
64 }