GCD
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4291 Accepted Submission(s): 1502
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.
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.
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
Recommend
wangye
前几天用容斥原理写过这题：
http://www.cnblogs.com/kuangbin/p/3269182.html
速度比较慢。
用莫比乌斯反演快很多。
莫比乌斯反演资料：
http://wenku.baidu.com/view/542961fdba0d4a7302763ad5.html
这题求[1,n],[1,m]gcd为k的对数。而且没有顺序。
转化之后就是[1,n/k],[1,m/k]之间互质的数的个数。
用莫比乌斯反演就很容易求了。
为了去除重复的，去掉一部分就好了；
这题求的时候还可以分段进行优化的。
具体看我的下一篇博客吧！
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013/8/21 19:32:35 4 File Name :F:2013ACM练习专题学习数学莫比乌斯反演HDU1695GCD.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 const int MAXN = 1000000; 21 bool check[MAXN+10]; 22 int prime[MAXN+10]; 23 int mu[MAXN+10]; 24 void Moblus() 25 { 26 memset(check,false,sizeof(check)); 27 mu[1] = 1; 28 int tot = 0; 29 for(int i = 2; i <= MAXN; i++) 30 { 31 if( !check[i] ) 32 { 33 prime[tot++] = i; 34 mu[i] = -1; 35 } 36 for(int j = 0; j < tot; j++) 37 { 38 if(i * prime[j] > MAXN) break; 39 check[i * prime[j]] = true; 40 if( i % prime[j] == 0) 41 { 42 mu[i * prime[j]] = 0; 43 break; 44 } 45 else 46 { 47 mu[i * prime[j]] = -mu[i]; 48 } 49 } 50 } 51 } 52 int main() 53 { 54 //freopen("in.txt","r",stdin); 55 //freopen("out.txt","w",stdout); 56 int T; 57 int a,b,c,d,k; 58 Moblus(); 59 scanf("%d",&T); 60 int iCase = 0; 61 while(T--) 62 { 63 iCase++; 64 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); 65 if(k == 0) 66 { 67 printf("Case %d: 0 ",iCase); 68 continue; 69 } 70 b /= k; 71 d /= k; 72 if(b > d)swap(b,d); 73 long long ans1 = 0; 74 for(int i = 1; i <= b;i++) 75 ans1 += (long long)mu[i]*(b/i)*(d/i); 76 long long ans2 = 0; 77 for(int i = 1;i <= b;i++) 78 ans2 += (long long)mu[i]*(b/i)*(b/i); 79 ans1 -= ans2/2; 80 printf("Case %d: %I64d ",iCase,ans1); 81 } 82 return 0; 83 }