zoukankan      html  css  js  c++  java
  • HDU1695 莫比乌斯反演

    GCD

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 15981    Accepted Submission(s): 6144


    Problem 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).
     
    Source
    题意:gcd=k的对数
    思路:倍数莫比乌斯反演。
    代码:
     1 #include<bits/stdc++.h>
     2 #define  ll long long
     3 using namespace std;
     4 const int N = 2e5 + 5;
     5 int t;
     6 //线性筛法求莫比乌斯函数
     7 bool vis[N + 10];
     8 int pri[N + 10];
     9 int mu[N + 10];
    10 int sum[N];
    11 int a,b;
    12 void mus() {
    13     memset(vis, 0, sizeof(vis));
    14     mu[1] = 1;
    15     int tot = 0;
    16     for (int i = 2; i < N; i++) {
    17         if (!vis[i]) {
    18             pri[tot++] = i;
    19             mu[i] = -1;
    20         }
    21         for (int j = 0; j < tot && i * pri[j] < N; j++) {
    22             vis[i * pri[j]] = 1;
    23             if (i % pri[j] == 0) {
    24                 mu[i * pri[j]] = 0;
    25                 break;
    26             }
    27             else  mu[i * pri[j]] = -mu[i];
    28         }
    29     }
    30     sum[1]=1;
    31     for(int i=2;i<N;i++) sum[i]=sum[i-1]+mu[i];
    32 }
    33 int n,m,k;
    34 ll cal(int x,int y){//求[1,x],[1,y]内互质的数对
    35     int ma=min(x,y);
    36     ll ans=0;
    37     for(int i=1,j;i<=ma;i=j+1){
    38         j=min(x/(x/i),y/(y/i));
    39         if(j>=ma) j=ma;
    40         ans+=1ll*(sum[j]-sum[i-1])*(x/i)*(y/i);
    41     }
    42     return ans;
    43 }
    44 int main() {
    45     mus();
    46     scanf("%d",&t);
    47     for(int i=1;i<=t;i++){
    48         scanf("%d%d%d%d%d",&a,&n,&b,&m,&k);
    49         ll ans,res;
    50         if(!k) ans=0;
    51         else
    52         {
    53             if(n<m) swap(n,m);
    54             res=cal(m/k,m/k);//这部分除1外,算了两次
    55             ans=cal(n/k,m/k);//这部分为总的个数
    56             ans=ans-res+(res+1)/2;//最后结果
    57         }
    58         printf("Case %d: %lld
    ",i,ans);
    59     }
    60     return 0;
    61 }
  • 相关阅读:
    交通综合改造工程EPC总承包项目
    二三维一体化地理信息平台
    NetCore3.1升级到Net5.0序列化方法过时问题
    windows server2012部署.net core IIS,页面报503,应用程序池自动停止。。。
    NetCore使用NPOI导入Word中的图片信息
    NetCore 使用 iTextSharp 读取 PDF 中的文字信息
    NetCore 在 Docker中文件路径找不到的问题
    Vue中数组list直接push的是对象而不是追加数据的问题
    netcore3.1增加阿里云OSS云存储服务
    Centos中Docker容器中程序访问宿主机Redis和Mysql
  • 原文地址:https://www.cnblogs.com/mj-liylho/p/9572455.html
Copyright © 2011-2022 走看看