求得n的因数后,简单容斥
Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1798 Accepted Submission(s): 685
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}.
/* *********************************************** Author :CKboss Created Time :2015年03月27日 星期五 17时52分24秒 File Name :HDOJ4135.cpp ************************************************ */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> using namespace std; typedef long long int LL; LL A,B,N; vector<LL> pr; void getPr() { LL TN=N,now=2; pr.clear(); while(TN!=1) { if(TN%now==0) { pr.push_back(now); while(TN%now==0) TN/=now; } now++; if(now*now>TN) break; } if(TN!=1) pr.push_back(TN); } LL RET,RET1,RET2; void dfs(int x,LL n,int mark) { for(int i=x,sz=pr.size();i<sz;i++) { RET+=mark*(n/pr[i]); dfs(i+1,n/pr[i],-mark); } } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int T_T,cas=1; cin>>T_T; while(T_T--) { cin>>A>>B>>N; getPr(); A--; RET=A; dfs(0,A,-1); RET1=RET; RET=B; dfs(0,B,-1); RET2=RET; cout<<"Case #"<<cas++<<": "<<RET2-RET1<<endl; } return 0; }