HDU 5597 GTW likes function 打表

最新推荐文章于 2025-09-07 00:18:35 发布

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最新推荐文章于 2025-09-07 00:18:35 发布

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文章标签： php 数据结构与算法

原文链接：http://www.cnblogs.com/qscqesze/p/5042003.html

本文解析了HDU 5596 GTW likes function题目，通过观察发现函数f(x)实际上等于x+1，进而简化为求解Euler totient function φ(n+x+1)，并提供了完整的C++实现代码。

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

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;
}

转载于:https://www.cnblogs.com/qscqesze/p/5042003.html