GTW likes function

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5596

Description

Now you are given two definitions as follows.

f(x)=∑xk=0(−1)k22x−2kCk2x−k+1,f0(x)=f(x),fn(x)=f(fn−1(x))(n≥1)

Note that φ(n) means Euler’s totient function.(φ(n)is an arithmetic function that counts the positive integers less than or equal to n that are relatively prime to n.)

For each test case, GTW has two positive integers — n and x, and he wants to know the value of the function φ(fn(x)).

Input

There is more than one case in the input file. The number of test cases is no more than 100. Process to the end of the file.

Each line of the input file indicates a test case, containing two integers, n and x, whose meanings are given above. (1≤n,x≤1012)

Output

In each line of the output file, there should be exactly one number, indicating the value of the function φ(fn(x)) of the test case respectively.

Sample Input

1 1

2 1

3 2

Sample Output

2

2

2

Hint

题意

题解：

一切反动派都是纸老虎

打表之后很容易发现，f(x) = x+1

于是这道题就很蠢了，直接输出phi(n+x+1)就好了

代码

#include<iostream>
#include<stdio.h>
using namespace std;


long long phi(long long n)
{
    long long tmp=n;
    for(long long i=2;i*i<=n;i++)
        if(n%i==0)
        {
            tmp/=i;tmp*=i-1;
            while(n%i==0)n/=i;
        }
    if(n!=1)tmp/=n,tmp*=n-1;
    return tmp;
}

int main()
{
    long long n,x;
    while(scanf("%I64d%I64d",&n,&x)!=EOF)
        printf("%I64d\n",phi(x+n+1));
    return 0;
}