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

2008 “Sunline Cup” National Invitational Contest

题意：

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\n",i,ans);
59     }
60     return 0;
61 }