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GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2332    Accepted Submission(s): 1185 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 3 1 1 10 2 10000 72   Sample Output 1 6 260   Source ECJTU 2009 Spring Contest   Recommend lcy   |   We have carefully selected several similar problems for you:   1695  1787  3501  2824  2582   

给定N,M求gcd(i,N)>=M的i的个数

我们可以分解N=a*b, i=a*d(b>=d 且b，d互质)，那么我们要求的就是a》=m的时候d的个数（b随a而确定）

由于b>=d且b，d互质，所以这个数目就是φ（b）-1  

但是，如果对于每个a枚举b，铁定超时。（仍然O((N-M)*sqrt(N))的复杂度）

但是如果单纯这样全部枚举的话依旧会超时，所以我们要想一个办法去优化它。

我们可以折半枚举，这里的折半并不是二分的意思。

我们先看，我们枚举时，当i<sqrt(n),假设a=n / i， 当i>sqrt(n)之后 有b=n/i，我们观察到当n%i==0时，会出现一种情况，就是a*b==n。所以我们就可以只需要枚举sqrt(n)种情况，然后和它对应的情况就是 n/i。

我们这种枚举时间会快非常多

#include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <map> #include <algorithm> using namespace std; #define pi acos(-1) #define endl '\n' #define srand() srand(time(0)); #define me(x,y) memset(x,y,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); cin.tie(0); #define FOR(x,n,i) for(int i=x;i<=n;i++) #define FOr(x,n,i) for(int i=x;i<n;i++) #define W while #define sgn(x) ((x) < 0 ? -1 : (x) > 0) #define bug printf("***********\n"); typedef long long LL; const int INF=0x3f3f3f3f; const LL LINF=0x3f3f3f3f3f3f3f3fLL; const int dx[]={-1,0,1,0,1,-1,-1,1}; const int dy[]={0,1,0,-1,-1,1,-1,1}; const int maxn=1000; const int maxx=1e7+100; const double EPS=1e-7; const int mod=998244353; template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} inline LL Scan() { LL Res=0,ch,Flag=0; if((ch=getchar())=='-')Flag=1; else if(ch>='0' && ch<='9')Res=ch-'0'; while((ch=getchar())>='0'&&ch<='9')Res=Res*10+ch-'0'; return Flag ? -Res : Res; } //freopen( "in.txt" , "r" , stdin ); //freopen( "data.out" , "w" , stdout ); //cerr << "run time is " << clock() << endl; int solve(int n) { int ans=n; if(n%2==0)//只搜n/2次，不然可能超一组 （也可能不超） { while(n%2==0) n/=2; ans=ans/2; } for(long long i=3;i*i<=n;i+=2) //每找到一个就更改上界，这个优化很多，所以直接copy刘汝佳的代码会超时 { if(n%i==0) { while(n%i==0) n/=i; ans=ans/i*(i-1); } } if(n>1) ans=ans/n*(n-1); return ans; } int main() { int t; t=Scan(); while(t--) { int n,m; n=Scan();m=Scan(); int ans=0; for(int i=1;i*i<=n;i++) { if(n%i==0) { if(i>=m) ans+=solve(n/i); if(n/i>=m&&i*i!=n) ans+=solve(i); } } cout<<ans<<endl; } }