作者:覃思慧_419 | 来源:互联网 | 2023-10-13 10:43
本文由编程笔记#小编为大家整理,主要介绍了GCD(欧拉函数)相关的知识,希望对你有一定的参考价值。
GCD
http://acm.hdu.edu.cn/showproblem.php?pid=2588
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3782 Accepted Submission(s): 2053
Problem Description
The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.
Input
The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.
Output
For each test case,output the answer on a single line.
Sample Input
Sample Output
1 #include
2 using namespace std;
3 #define lson l,mid,rt<<1
4 #define rson mid+1,r,rt<<1|1
5 #define IT set::iterator
6 #define pb push_back
7 #define eb emplace_back
8 #define maxn 200005
9 #define eps 1e-6
10 #define PI acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pairpll;
15 typedef pairint> pli;
16 typedef pairint,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long MOD=1000000007;
19 const double oula=0.57721566490153286060651209;
20 const int INF=0x3f3f3f3f;
21 using namespace std;
22
23
24 ll Euler(ll n){
25 ll ans=n;
26 for(ll i=2;i*i<=n;i++){
27 if(n%i==0) ans=ans/i*(i-1);
28 while(n%i==0) n/=i;
29 }
30 if(n>1) ans=ans/n*(n-1);
31 return ans;
32 }
33
34 int main(){
35 std::ios::sync_with_stdio(false);
36 int t;
37 cin>>t;
38 while(t--){
39 ll n,m;
40 cin>>n>>m;
41 ll ans=0;
42 for(int i=1;i*i<=n;i++){
43 if(n%i==0){
44 if(i>=m) ans+=Euler(n/i);
45 if(i*i!=n&&n/i>=m) ans+=Euler(i);
46 }
47 }
48 cout<endl;
49 }
50 }
View Code