GCD
Time Limit: 6000/3000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 5106 Accepted
Submission(s): 1833
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).
Time Limit: 6000/3000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 5106 Accepted
Submission(s): 1833
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).
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <math.h>
5 #include <algorithm>
6 #include <vector>
7 using namespace std;
8 vector<int>q[110000];
9 long long a[110000]={0};
10 int bb;
11 void init()
12 {
13 int i,j;
14 for(i=0; i<110000; i++)a[i]=i,q[i].clear();
15 for(i=2; i<110000; i+=2)
16 a[i]/=2,q[i].push_back(2);
17 for(i=3; i<110000; i+=2)
18 if(a[i]==i)
19 for(j=i; j<110000; j+=i)
20 a[j]=a[j]/i*(i-1),q[j].push_back(i);
21 for(i=1; i<110000; i++)
22 a[i]+=a[i-1];
23 }
24 int fun(int x,int y)
25 {
26 int i,cnt=0;
27 int sum=1;
28 for(i=0;i<q[y].size();i++)
29 {
30 if(x&(1<<i))
31 {
32 sum*=q[y][i];
33 cnt++;
34 }
35 }
36 if(cnt&1)
37 return bb/sum;
38 else return -(bb/sum);
39 }
40 long long work(int x)
41 {
42 int i;
43 long long sum=0;
44 for(i=1;i<(1<<q[x].size());i++)
45 {
46 sum+=fun(i,x);
47 }
48 return bb-sum;
49 }
50 int main()
51 {
52 init();
53 int i,t,j,aa,c,d,k;
54 long long ans;
55 scanf("%d",&t);
56 for(i=1; i<=t; i++)
57 {
58 scanf("%d%d%d%d%d",&aa,&bb,&c,&d,&k);
59 if(bb>d)swap(bb,d);
60 if(k)
61 bb/=k,d/=k;
62 else
63 {
64 printf("Case %d: %d\n",i,0);
65 continue;
66 }
67 ans=a[bb];
68 for(j=bb+1; j<=d; j++)
69 {
70 ans+=work(j);
71 }
72 printf("Case %d: %I64d\n",i,ans);
73 }
74 }
GCD hdu1695容斥原理,布布扣,bubuko.com
GCD hdu1695容斥原理
原文：http://www.cnblogs.com/ERKE/p/3682068.html
