hdoj5597GTW likes function【找规律+欧拉函数】

GTW likes function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 241    Accepted Submission(s): 132


Problem 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

 

Source

BestCoder Round #66 (div.2)

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
long long Euler(long long n){
long long ret=1,i;
for(i=2;i*i<=n;++i){
if(n%i==0){
n/=i;ret*=i-1;
while(n%i==0){
n/=i;ret*=i;
}
}
}
if(n>1)ret*=n-1;
return ret;
}
int main()
{
long long n,x;
while(scanf("%lld%lld",&n,&x)!=EOF){
printf("%lld\n",Euler(n+x+1));
}
return 0;
}