题目链接:http://hdu.hustoj.com/showproblem.php?pid=2588
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.
(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
3
1 1
10 2
10000 72
Sample Output
1
6
260
题目大意:
给定n和m,求有多少x满足gcd(x,n)>=m。
题目思路:
假设一个gcd(x,n)=s
那么s*a = x;
s*b = n;
则a<=b(因为x<=n)且gcd(a,b)=1;
发现什么没有,这就是欧拉函数,求出小于等于b且与b互质的a的个数就是x的数量。
所以我们就可以枚举s,要求s>=m。因为n比较大,所以枚举到sqrt(n),另一个直接算。
欧拉函数代码可看:
代码:
#include<cstdio> #include<cmath> #include<cstring> #include<string> #include<cstdlib> #include<algorithm> #include<iostream> #include<queue> #include<stack> #include<map> using namespace std; #define FOU(i,x,y) for(int i=x;i<=y;i++) #define FOD(i,x,y) for(int i=x;i>=y;i--) #define MEM(aA,val) memset(a,val,sizeof(a)) #define PI acos(-1.0) const double EXP = 1e-9; typedef long long ll; typedef unsigned long long ull; const int INF = 0x3f3f3f3f; const ll MINF = 0x3f3f3f3f3f3f3f3f; const double DINF = 0xffffffffffff; const int mod = 1e9+7; int phi(int x){ int ans=x; for(int i=2;i*i<=x;i++){ if(x%i == 0){ ans = ans/i * (i-1); while(x % i==0) x/=i; } } if(x>1) ans = ans/x*(x-1); return ans; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); std::ios::sync_with_stdio(false); int t,n,m; cin>>t; while(t--) { cin>>n>>m; ll ans=0; for(int i=1;i*i<=n;i++) { if(n%i==0) { if(i>=m) ans+=phi(n/i); if(i*i!=n) { if(n/i>=m) ans+=phi(i); } } } cout<<ans<<endl; } return 0; }