hdu 1695 莫比乌斯反演

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8482    Accepted Submission(s): 3152

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

21 3 1 5 11 11014 1 14409 9

 

Sample Output

Case 1: 9Case 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).
莫比乌斯反演资料：
http://wenku.baidu.com/view/542961fdba0d4a7302763ad5.html
http://baike.baidu.com/link?url=1qQ-hkgOwDJAH4xyRcEQVoOTmHbiRCyZZ-hEJxRBQO8G0OurXNr6Rh6pYj9fhySI0MY2RKpcaSPV9X75mQv0hK
#include<iostream>#include<cstring>#include<cstdlib>#include<cstdio>#include<cmath>#include<set>#include <queue>#include<algorithm>const double PI = acos(-1.0);using namespace std;int mu[100005], vis[100005], prime[100005];void Mobuls(){memset(vis, 0 ,sizeof(vis));mu[1] = 1;int top = 0;for(int i = 2; i <= 100000; ++ i){if(!vis[i]){prime[top ++ ] = i;mu[i] = -1;}for(int j = 0; j < top; ++ j){if(i * prime[j] > 100000)break;vis[i * prime[j]] = 1;if(i % prime[j] == 0){mu[i * prime[j]] = 0;break;}elsemu[i * prime[j]] = -mu[i];}}}int main(){int n, a, b, c, d, k, i, j;int t = 1;scanf("%d", &n);Mobuls();while(n --){scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);if(k == 0){printf("Case %d: 0\n", t ++);continue;}b /= k;d /= k;long long ans1 = 0, ans2 = 0;for(i = 1; i <= max(b, d); ++ i)ans1 += (long long) mu[i] * (b / i) * (d / i);for(i = 1; i <= min(b, d); ++ i)ans2 += (long long) mu[i] * (min(b, d) / i) * (min(b, d) / i);printf("Case %d: %lld\n",t ++, ans1 - ans2 / 2);}}