zoukankan      html  css  js  c++  java
  • HDU 1787 GCD Again(欧拉函数,水题)

    GCD Again

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1997    Accepted Submission(s): 772


    Problem Description
    Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
    No? Oh, you must do this when you want to become a "Big Cattle".
    Now you will find that this problem is so familiar:
    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 I am considering a little more difficult problem: 
    Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
    This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
    Good Luck!
     
    Input
    Input contains multiple test cases. Each test case contains an integers N (1<N<100000000). A test case containing 0 terminates the input and this test case is not to be processed.
     
    Output
    For each integers N you should output the number of integers M in one line, and with one line of output for each line in input. 
     
    Sample Input
    2 4 0
     
    Sample Output
    0 1
     
    Author
    lcy
     
    Source
     
    Recommend
    lcy

    用来试验下模板。

    求欧拉函数就可以了

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    long long eular(long long n)
    {
        long long ans = n;
        for(int i = 2;i*i <= n;i++)
        {
            if(n % i == 0)
            {
                ans -= ans/i;
                while(n % i == 0)
                    n /= i;
            }
        }
        if(n > 1)ans -= ans/n;
        return ans;
    }
    
    int main()
    {
        int n;
        while(scanf("%d",&n) == 1 && n)
        {
            int ret = eular(n);
            printf("%d
    ",n-ret-1);
        }
        return 0;
    }
  • 相关阅读:
    聊一聊Java泛型的擦除
    微信退款通知信息解密
    Spring Boot 初识
    shiro初识
    Redis 初探
    Java Json库的选取准则
    JAVA 几款Json library的比较
    FUSE简介
    Lab 2 源码分析
    Lab2
  • 原文地址:https://www.cnblogs.com/kuangbin/p/3205712.html
Copyright © 2011-2022 走看看