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
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 ",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 ",solve(B,n)-solve(A-1,n)); 51 } 52 return 0; 53 }