zoukankan      html  css  js  c++  java
  • SPOJ PGCD 4491. Primes in GCD Table && BZOJ 2820 YY的GCD (莫比乌斯反演)

    4491. Primes in GCD Table

    Problem code: PGCD


    Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has created a GCD (greatest common divisor) table! So he now has a table (of height a and width b), indexed from (1,1) to (a,b), and with the value of field (i,j) equal to gcd(i,j). He wants to know how many times he has used prime numbers when writing the table.

    Input

    First, t ≤ 10, the number of test cases. Each test case consists of two integers, 1 ≤ a,b < 107.

    Output

    For each test case write one number - the number of prime numbers Johnny wrote in that test case.

    Example

    Input:
    2
    10 10
    100 100
    Output:
    30
    2791

    题解参考:

    http://quartergeek.com/eight-gcd-problems/

    ans = sigma(p, sigma(d, μ(d) * (n/pd) * (m/pd)))
    
    Let s = pd, then
    
    ans = sigma(s, sigma(p, μ(s/p) * (n/s) * (m/s)))
        = sigma(s, (n/s) * (m/s) * sigma(p, μ(s/p)))
    
    Let g(x) = sigma(p, μ(x/p)), then
    
    ans = sigma(s, (n/s) * (m/s) * g(s))

    如果我们能预处理g(x)的话就能和前面一样分块搞了。这个时候我们多么希望g(x)μ(x)一样是积性函数。看完题解后,发现有一个不是积性函数,胜似积性函数的性质。由于题解没有给证明,所以就意淫了一个证明。

    考虑质数p'g(p'x) = sigma(p | p'x, μ(p'x/p))

    • x % p' == 0,那么考虑sigma中的变量p的所有取值,它和g(x)p是相同的。而μ(x)这个函数,如果x有平方因子的话就等于0,因此当p != p'μ(p'x/p) = 0,当p == p'是,μ(p'x/p) = μ(x)。所以g(p'x) = μ(x)
    • x % p' != 0,同样考虑p,会发现它的取值只比g(x)中的p多出一个p'。同理按照p是否等于p'讨论,可以得到g(p'x) = -g(x) + μ(x)

    因此g(x)可以在线性筛素数的时候算出。剩下的就是前缀和、分块了。

     1 /* ***********************************************
     2 Author        :kuangbin
     3 Created Time  :2013-10-19 22:01:05
     4 File Name     :E:2013ACM专题学习数学莫比乌斯反演SPOJ_PGCD.cpp
     5 ************************************************ */
     6 
     7 #include <stdio.h>
     8 #include <string.h>
     9 #include <iostream>
    10 #include <algorithm>
    11 #include <vector>
    12 #include <queue>
    13 #include <set>
    14 #include <map>
    15 #include <string>
    16 #include <math.h>
    17 #include <stdlib.h>
    18 #include <time.h>
    19 using namespace std;
    20 
    21 const int MAXN = 10000000;
    22 bool check[MAXN+10];
    23 int prime[MAXN+10];
    24 int mu[MAXN+10];
    25 int g[MAXN+10];
    26 int sum[MAXN+10];
    27 void Moblus()
    28 {
    29     memset(check,false,sizeof(check));
    30     mu[1] = 1;
    31     int tot = 0;
    32     for(int i = 2; i <= MAXN; i++)
    33     {
    34         if(!check[i])
    35         {
    36             prime[tot++] = i;
    37             mu[i] = -1;
    38             g[i] = 1;
    39         }
    40         for(int j = 0;j < tot;j++)
    41         {
    42             if(i * prime[j] > MAXN)break;
    43             check[i*prime[j]] = true;
    44             if(i % prime[j] == 0)
    45             {
    46                 mu[i * prime[j]] = 0;
    47                 g[i * prime[j]] = mu[i];
    48                 break;
    49             }
    50             else
    51             {
    52                 mu[i * prime[j]] = -mu[i];
    53                 g[i * prime[j]] = -g[i] + mu[i];
    54             }
    55         }
    56     }
    57     sum[0] = 0;
    58     for(int i = 1;i <= MAXN;i++)
    59         sum[i] = sum[i-1] + g[i];
    60 }
    61 int main()
    62 {
    63     //freopen("in.txt","r",stdin);
    64     //freopen("out.txt","w",stdout);
    65     Moblus();
    66     int T;
    67     int n,m;
    68     scanf("%d",&T);
    69     while(T--)
    70     {
    71         scanf("%d%d",&n,&m);
    72         if(n > m)swap(n,m);
    73         long long ans = 0;
    74         int last = 0;
    75         for(int i = 1;i <= n;i = last+1)
    76         {
    77             last = min(n/(n/i),m/(m/i));
    78             ans += (long long)(sum[last] - sum[i-1])*(n/i)*(m/i);
    79         }
    80         printf("%lld
    ",ans);
    81     }
    82     return 0;
    83 }

  • 相关阅读:
    腾讯社招 —— 腾讯游戏后端工程师-电话面试
    乐刷科技-Java工程师社招面试
    平安人寿保险-深圳Java开发工程师社招面试
    字节跳动-后端工程师社招视频一面
    Markdown的常用使用语法
    oracle 表空间大小
    mysql、sqlserver、oracle获取最后一条数据
    mycat 学习
    oracle sql中特殊字符 & 的处理
    map循环
  • 原文地址:https://www.cnblogs.com/kuangbin/p/3378531.html
Copyright © 2011-2022 走看看