HDU 2588 GCD

GCD

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

思路：

\(ans=\sum_{i \in \{x|n\%x=0\}}\phi(\frac{n}{i})\)


证明：

设\(M=it\)

\(\Rightarrow t \leq \frac{n}{i} \And (\frac{n}{i},t)=1\)

\(So\ \ t\)的可行数量为\(\phi(\frac{n}{i})\)

答案为\(ans=\sum_{i \in \{x|n\%x=0\}}\phi(\frac{n}{i})\)

\(\mathfrak{Talk\ is\ cheap,show\ you\ the\ code.}\)

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
# define Type template<typename T>
# define read read1<int>()
Type inline T read1()
{
	T t=0;
	bool ty=0;
	char k;
	do k=getchar(),(k=='-')&&(ty=1);while('0'>k||k>'9');
	do t=(t<<3)+(t<<1)+(k^'0'),k=getchar();while('0'<=k&&k<='9');
	return ty?-t:t;
}
# define int long long
# define fre(k) freopen(k".in","r",stdin);freopen(k".out","w",stdout)
int work(int n)
{
    int tn=n;
    for(int i=2;i*i<=n;++i)
        if(!(n%i))
        {
            while(!(n%i))n/=i;
            tn=tn/i*(i-1);
        }
    if(n!=1)tn=tn/n*(n-1);
    return tn;
}
signed main()
{
    for(int T=read;T--;)
    {
        int s=read,m=read,ans=0;
        for(int i=1;i*i<=s;++i)
            if(!(s%i))
                if(i*i==s&&i>=m)ans+=work(i);
                else
                {
                    if(i>=m)ans+=work(s/i);
                    if(s/i>=m)ans+=work(i);
                }
        printf("%lld\n",ans);
    }
	return 0;
}