zoukankan      html  css  js  c++  java
  • HDU 1695 GCD(求两区间的互质数对+容斥原理)

    Description

    Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs. 
    Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same. 

    Yoiu can assume that a = c = 1 in all test cases.
     

    Input

    The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases. 
    Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above. 
     

    Output

    For each test case, print the number of choices. Use the format in the example. 
     

    Sample Input

    2 1 3 1 5 1 1 11014 1 14409 9
     

    Sample Output

    Case 1: 9 Case 2: 736427

    Hint

    For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5). 

    b,d除于k,转化为求互质对
    用容斥原理求出所有不互质的数对数,再用整数减去!

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <vector>
    #include <cstring>
    
    using namespace std;
    
    #define LL long long
    #define N 111111
    
    int a,b,c,d,k;
    LL ans;
    vector<int> prime[N];
    bool vis[N];
    
    void init(){
        memset(vis,false,sizeof vis);
        for(int i=0;i<=N;i++) prime[i].clear();
        for(int i=2;i<=N;i+=2) prime[i].push_back(2);//这样快很多
        for(int i=3;i<=N;i+=2) if(!vis[i]){
            for(int j=i;j<=N;j+=i){
                prime[j].push_back(i);vis[j]=true;
            }
        }
    }
    
    void fun(int x,LL y,int z){
        LL v = 1,cnt=0;
        for(int i=0;i<prime[x].size();i++){
            if(1<<i&y){
                v*=prime[x][i];
                cnt++;
            }
        }
        if(cnt&1) ans-=z/v;
        else ans+=z/v;
    }
    
    
    int main()
    {
        init();int _,o;scanf("%d",&_);o=_;
        while(_--){
            printf("Case %d: ",o-_);ans=0;
            scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
            if(!k) {puts("0");continue;}
            b/=k,d/=k;int z;if(d<b) z=d,d=b,b=z;
            for(int i=1;i<=d;i++){
                int k = min(i,b);ans+=k;//保证不重复
                for(LL j=1;j<(1<<prime[i].size());j++) fun(i,j,k);
            }
            printf("%lld
    ",ans);
        }
        return 0;
    }
    

      



  • 相关阅读:
    利用Airtest开发爬虫
    添加mitmproxy证书为安卓系统证书(模拟器亦可)
    mac下 安卓模拟器抓包推荐
    mac上appium连接ios
    HomeBrew和HomeBrew Cask的安装和使用
    NOIP2020 游记
    CSP2020 游记
    随机数之 xorShift128Plus 板子
    最大单词长度乘积(力扣第318题)
    两个整数之和(力扣第371题)
  • 原文地址:https://www.cnblogs.com/BugClearlove/p/4705465.html
Copyright © 2011-2022 走看看