杭电5597

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GTW likes function

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 205 Accepted Submission(s): 112

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 12 13 2

Sample Output

Source

BestCoder Round #66 (div.2)

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下次就不证明了，听说大牛都是打表得到的规律，fn(x)=x+n+1;这个但是很容易得到：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;long long ac(long long n){long long res=n;long long a=n;for(long long i=2;i*i<=a;i++){if(a%i==0)res=res/i*(i-1);while(a%i==0)a/=i;}if(a>1)res=res/a*(a-1);return res;}int main(){long long m,n;while(scanf("%I64d%I64d",&m,&n)!=EOF){long long ans=ac(m+n+1);printf("%I64d\n",ans);}}

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