• [POI 2007]ZAP-Queries


    Description

    Byteasar the Cryptographer works on breaking the code of BSA (Byteotian Security Agency). He has alreadyfound out that whilst deciphering a message he will have to answer multiple queries of the form"for givenintegers $a$, $b$ and $d$, find the number of integer pairs $(x,y)$ satisfying the following conditions:

    $1le xle a$,$1le yle b$,$gcd(x,y)=d$, where $gcd(x,y)$ is the greatest common divisor of $x$ and $y$".

    Byteasar would like to automate his work, so he has asked for your help.

    TaskWrite a programme which:

    reads from the standard input a list of queries, which the Byteasar has to give answer to, calculates answers to the queries, writes the outcome to the standard output.

    FGD正在破解一段密码,他需要回答很多类似的问题:对于给定的整数a,b和d,有多少正整数对x,y,满足x<=a,y<=b,并且gcd(x,y)=d。作为FGD的同学,FGD希望得到你的帮助。

    Input

    The first line of the standard input contains one integer $n$ ($1le nle 50 000$),denoting the number of queries.

    The following $n$ lines contain three integers each: $a$, $b$ and $d$($1le dle a,ble 50 000$), separated by single spaces.

    Each triplet denotes a single query.

    Output

    Your programme should write $n$ lines to the standard output. The $i$'th line should contain a single integer: theanswer to the $i$'th query from the standard input.

    Sample Input

    2
    4 5 2
    6 4 3

    Sample Output

    3
    2

    题解

    双倍经验。去掉容斥就好了,分析见超链接。

     1 //It is made by Awson on 2018.1.21
     2 #include <set>
     3 #include <map>
     4 #include <cmath>
     5 #include <ctime>
     6 #include <queue>
     7 #include <stack>
     8 #include <cstdio>
     9 #include <string>
    10 #include <vector>
    11 #include <cstdlib>
    12 #include <cstring>
    13 #include <iostream>
    14 #include <algorithm>
    15 #define LL long long
    16 #define Abs(a) ((a) < 0 ? (-(a)) : (a))
    17 #define Max(a, b) ((a) > (b) ? (a) : (b))
    18 #define Min(a, b) ((a) < (b) ? (a) : (b))
    19 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
    20 #define writeln(x) (write(x), putchar('
    '))
    21 #define lowbit(x) ((x)&(-(x)))
    22 using namespace std;
    23 const int N = 50000;
    24 void read(int &x) {
    25     char ch; bool flag = 0;
    26     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    27     for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    28     x *= 1-2*flag;
    29 }
    30 void write(LL x) {
    31     if (x > 9) write(x/10);
    32     putchar(x%10+48);
    33 }
    34 
    35 int a, b, k, mu[N+5];
    36 
    37 void get_mu() {
    38     int isprime[N+5], prime[N+5], tot = 0;
    39     memset(isprime, 1, sizeof(isprime)); isprime[1] = 0, mu[1] = 1;
    40     for (int i = 2; i <= N; i++) {
    41     if (isprime[i]) mu[i] = -1, prime[++tot] = i;
    42     for (int j = 1; j <= tot && i*prime[j] <= N; j++) {
    43         isprime[i*prime[j]] = 0;
    44         if (i%prime[j]) mu[i*prime[j]] = -mu[i];
    45         else {mu[i*prime[j]] = 0; break; }
    46     }
    47     mu[i] += mu[i-1];
    48     }
    49 }
    50 LL cal(int a, int b) {
    51     if (a > b) Swap(a, b); LL ans = 0;
    52     for (int i = 1, last; i <= a; i = last+1) {
    53     last = Min(a/(a/i), b/(b/i));
    54     ans += (LL)(mu[last]-mu[i-1])*(a/i)*(b/i);
    55     }
    56     return ans;
    57 }
    58 void work() {
    59     read(a), read(b), read(k); writeln(cal(a/k, b/k));
    60 }
    61 int main() {
    62     int t; read(t); get_mu();
    63     while (t--) work();
    64     return 0;
    65 }
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  • 原文地址:https://www.cnblogs.com/NaVi-Awson/p/8325830.html
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