GCD
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7026 Accepted Submission(s):
2584
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： 给出a,b,c,d,k, 使得 a<=x<=b , c<=y<=d &&gcd(x,y)=k的数对（x,y）的对数。
限制： a=c=1 , 0<b,c<=100000 ;(n1,n2)和（n2 ,n1）算为同一种情况、
做这道题之前 ，先来了解一下，莫比乌斯函数：
知道，莫比乌斯函数，求其反演函数.....具体可以见上面的地址
对于这道题，我们可以这样分析：
1 设f(k)为gcd(x,y)=k的数对(x,y)的对数，我们要求的是f(1) 2 设F(k)为gcd(x,y)为k的倍数的数对(x,y)的对数，可以想到F(k)=floor(b/k)*floor(d/k)， F(k)=E[b/k]*[d/k]*mu[k]; 3 由莫比乌斯反演得： 4 令lim=min(b/k,d/k) 5 f(1)=mu[1]*F(1) + mu[2]*F[2] + ... + mu[lim]*F(lim) 6 因为(n1,n2)和(n2,n1)算为同一种情况，所以最后结果还要减掉重复的情况。
代码:
1 #include<cstdio>
2 #include<cstdlib>
3 #include<cstring>
4 #include<iostream>
5 #include<algorithm>
6 #define _llint long long
7 using namespace std;
8 const int maxn=100010;
9 bool jud[maxn];
10 int mu[maxn] , prime[maxn];
11
12 void Mobius(){
13 int cnt=0;
14 memset(jud , 0 , sizeof jud);
15 mu[1]=1; //d=1
16 for(int i=2 ; i<maxn ;i++){
17
18 if(!jud[i]){
19 mu[i]=-mu[1]; //1*p
20 prime[cnt++]=i;
21 }
22 for(int j=0 ; j<cnt&&i*prime[j]<maxn ; j++ ){
23
24 jud[i*prime[j]]=1;
25 if(i%prime[j]==0){
26 mu[i*prime[j]]=0 ; //²»ÊÇ»¥ÒìµÄËØÊý
27 break;
28 }else{
29 mu[i*prime[j]]= -1*mu[i];
30 }
31 }
32 }
33 return ;
34 }
35
36 _llint solve(int n, int m){
37
38 _llint res=0;
39 for(int i=1; i<=n ;i++){
40 res+=(_llint)(m/i)*(n/i)*mu[i];
41 }
42 return res;
43 }
44
45 int main()
46 {
47 int T,a,b,c,d,k;
48 Mobius();
49 scanf("%d",&T);
50 for(int i=1 ;i<=T ;i++){
51 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
52 if(0==k)
53 {
54 printf("Case %d: 0\n",i);
55 continue;
56 }
57 b/=k , d/=k;
58 if(b>d) swap( b , d );
59 _llint ans1=solve(b,d);
60 _llint ans2=solve(b,b);
61 printf("Case %d: %lld\n",i,ans1-ans2/2);
62 }
63 return 0;
64 }