Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2307 Accepted Submission(s): 861
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
题意：求A到B之间的数有多少个与n互质。
首先转化为（1---B）与n互质的个数减去（1--- A-1）与n互质的个数
然后就是求一个区间与n互质的个数了，注意如果是求（1---n）与n互质的个数，可以用欧拉函数，但是这里不是到n，所以无法用欧拉函数。
这里用到容斥原理，即将求互质个数转化为求不互质的个数，然后减一下搞定。
求互质个数的步骤：
1、先将n质因数分解
2、容斥原理模板求出不互质个数ans
3、总的个数减掉不互质个数就得到答案
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<set>
5 #include<vector>
6 using namespace std;
7 #define ll long long
8 #define N 1000000
9 ll A,B,n;
10 vector<ll> v;
11 ll solve(ll x,ll n)
12 {
13 v.clear();
14 for(ll i=2;i*i<=n;i++) //对n进行素数分解
15 {
16 if(n%i==0)
17 {
18 v.push_back(i);
19 while(n%i==0)
20 n/=i;
21 }
22 }
23 if(n>1) v.push_back(n);
24
25 ll ans=0;
26 for(ll i=1;i<( 1<<v.size() );i++)//用二进制来1,0来表示第几个素因子是否被用到,如m=3，三个因子是2,3,5，则i=3时二进制是011，表示第2、3个因子被用到
27 {
28 ll sum=0;
29 ll tmp=1;
30 for(ll j=0;j<v.size();j++)
31 {
32 if((1<<j)&i) //判断第几个因子目前被用到
33 {
34 tmp=tmp*v[j];
35 sum++;
36 }
37 }
38 if(sum&1) ans+=x/tmp;//容斥原理，奇加偶减
39 else ans-=x/tmp;
40 }
41 return x-ans;
42 }
43 int main()
44 {
45 int t;
46 int ac=0;
47 scanf("%d",&t);
48 while(t--)
49 {
50 scanf("%I64d%I64d%I64d",&A,&B,&n);
51 printf("Case #%d: ",++ac);
52 printf("%I64d\n",solve(B,n)-solve(A-1,n));
53 }
54 return 0;
55 }
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 #define ll long long
6 #define N 1000000
7 ll A,B,n;
8 ll fac[N];
9 ll solve(ll x,ll n)
10 {
11 ll num=0;
12 for(ll i=2;i*i<=n;i++)
13 {
14 if(n%i==0)
15 {
16 fac[num++]=i;
17 while(n%i==0)
18 n/=i;
19 }
20 }
21 if(n>1) fac[num++]=n;
22
23 ll ans=0;
24 for(ll i=1;i<(1<<num);i++)
25 {
26 ll sum=0;
27 ll tmp=1;
28 for(ll j=0;j<num;j++)
29 {
30 if((1<<j)&i)
31 {
32 tmp=tmp*fac[j];
33 sum++;
34 }
35 }
36 if(sum&1) ans+=x/tmp;
37 else ans-=x/tmp;
38 }
39 return x-ans;
40 }
41 int main()
42 {
43 int t;
44 int ac=0;
45 scanf("%d",&t);
46 while(t--)
47 {
48 scanf("%I64d%I64d%I64d",&A,&B,&n);
49 printf("Case #%d: ",++ac);
50 printf("%I64d\n",solve(B,n)-solve(A-1,n));
51 }
52 return 0;
53 }