zoukankan      html  css  js  c++  java
  • hdu1695 GCD(莫比乌斯反演)

    题意:求(1,b)区间和(1,d)区间里面gcd(x, y) = k的数的对数(1<=x<=b , 1<= y <= d)。

     

    知识点:

    莫比乌斯反演/*12*/

    线性筛求莫比乌斯反演函数:

    void Init()
    {
        memset(vis,0,sizeof(vis));
        mu[1] = 1;
        cnt = 0;
        for(int i=2; i<N; i++)
        {
            if(!vis[i])
            {
                prime[cnt++] = i;
                mu[i] = -1;
            }
            for(int j=0; j<cnt&&i*prime[j]<N; j++)
            {
                vis[i*prime[j]] = 1;
                if(i%prime[j]) mu[i*prime[j]] = -mu[i];
                else
                {
                    mu[i*prime[j]] = 0;
                    break;
                }
            }
        }
    }

     

    题解:

    转化题意就是[1,n/k],[1,m/k]之间互质的数的个数。

     

     

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    const int N=100000+10;
    int u[N],prime[N];
    bool vis[N];
    
    void init()
    {
        memset(vis,0,sizeof(vis));
        u[1] = 1;
        int cnt = 0;
        for(int i=2; i<N; i++)
        {
            if(!vis[i])
            {
                prime[cnt++] = i;
                u[i] = -1;
            }
            for(int j=0; j<cnt&&i*prime[j]<N; j++)
            {
                vis[i*prime[j]] = 1;
                if(i%prime[j]) u[i*prime[j]] = -u[i];
                else
                {
                    u[i*prime[j]] = 0;
                    break;
                }
            }
        }
    }
    int main()
    {
        init();
        int t;
        cin>>t;
        int a,b,c,d,k;
        for(int kase=1;kase<=t;kase++)
        {
            scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
            if(k==0)
            {
                printf("Case %d: 0
    ",kase);
                continue;
            }
            long long ans=0;
            int ma=max(b,d),mi=min(b,d);
            for(int i=k;i<=mi;i+=k)
            {
                ans+=(long long)u[i/k]*((ma/i)*2-(mi/i)+1)*(mi/i)/2;
            }
            printf("Case %d: %I64d
    ",kase,ans);
        }
        return 0;
    }

     

     

    GCD

    Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

    Submit Status

    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). 
  • 相关阅读:
    Mapreduce实例——MapReduce自定义输出格式
    Mapreduce实例——MapReduce自定义输入格式
    Mapreduce实例——ChainMapReduce
    Mapreduce实例——倒排索引
    Cache 和 Buffer 都是缓存,主要区别是什么?
    Pycharm 2019 破解激活方法
    Python 两个list合并成一个字典
    Python中常见字符串去除空格的方法总结
    操作的系统的PV操作
    信号与信号量的区别
  • 原文地址:https://www.cnblogs.com/shentr/p/5349736.html
Copyright © 2011-2022 走看看