（hdu2588）GCD（欧拉函数）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2320 Accepted Submission(s): 1174

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

题意：两个整数N，M，问有多少个X满足条件1<=X<=N且gcd(X,N)>=M

分析：可以直接用欧拉函数求解

#include<cstdio>#include<cmath>using namespace std;int Euler(int n)///欧拉函数求小于等于n的正整数中有多少个数与n互质{    int s=n,i;    for(i=2; i<=sqrt(n); i++)     if(n%i==0)        {            s=s/i*(i-1);            while(n%i==0) n/=i;        }    if(n>1) s=s/n*(n-1);    return s;}int main(){    int t,n,m;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        if(m==1) {printf("%d\n",n);continue;}        int num=0;        for(int i=2; i*i<=n; i++)        {            if(n%i==0) //i是n的因子            {                if(i>=m)                    num+=Euler(n/i);                if(i*i!=n)///避免重复计算n/i的欧拉函数所求值                {                    if(n/i>=m)                        num+=Euler(n/(n/i));                }            }        }        printf("%d\n",num+1);///+1是因为有一个满足条件的数是n本身    }    return 0;}

可以用容斥原理写？ 我试了，时间超限，用队列和二进制位实现容斥原理都会TLE。。 递归？
先把代码放上来吧。。

#include<cstdio>#include<algorithm>#include<vector>using namespace std;typedef long long LL;int n,m;LL gcd(LL a,LL b){    return b?gcd(b,a%b):a;}LL lcm(int m,int n){    return m/gcd(m,n)*n;}int solve(){    if(m==1) return n;    vector<int>g;    int i,j;    for(int i=1; i<=n; i++)    {        if(n%i==0)        {            if(i>=m)                g.push_back(i);            if(n/i>=m)                g.push_back(n/i);        }    }    sort(g.begin(),g.end());    int cnt=1;    for(i=1; i<g.size(); i++)    {        for(j=0; j<i; j++)            if(g[i]%g[j]==0)                break;        if(j==i)            g[cnt++]=g[i];    }    int sum=0;    for(int k=1; k<(1<<cnt); k++)    {        int mul=1,num=0;        for(int i=0; i<cnt; i++)        {            if(k&(1<<i))            {                num++;                mul=lcm(mul,g[i]);            }        }        if(num&1) sum+=n/mul;        else sum-=n/mul;    }    return sum;    /* LL sum=0,front=0;    LL k,que[N],t,tmp;    que[front++]=-1;    for(LL i=0; i<cnt; i++)    {        k=front;        for(LL j=0; j<k; j++)        {            tmp=abs(que[j]);            t=lcm(tmp,g[i]);            que[front++]=t/que[j]*abs(que[j])*(-1);        }    }    for(LL i=1; i<front; i++)    {        //printf("%d\n",que[i]);        sum+=n/que[i];    }    return sum;*/}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        printf("%d\n",solve());    }    return 0;}