hdu1695—GCD（求两个区间内互质的数的对数）

题目链接：传送门

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11844    Accepted Submission(s): 4465

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).
 

 

解题思路：求两个区间内gcd(x,y) = k,那么x/k与y/k互质，即求区间[1,b/k]与[1,d/k]内有多少对数互质,可以先求一个数与某个区间互质的数的个数。

#include <iostream>#include <cstring>#include <cstdio>#include <vector>using namespace std;typedef long long ll;const int N = 100300;const int INF = 0x3f3f3f3f;int solve(int n,int r){    vector<int>p;    for(int i=2; i*i<=n; ++i)       if(n%i == 0){           p.push_back (i);           while(n%i == 0)                n /= i;       }    if(n > 1)        p.push_back(n);    int sz = p.size();    int sum = 0;    for(int i=1; i<(1<<sz); ++i){       int mult = 1,bits = 0;       for (int j=0; j<sz; ++j)           if (i&(1<<j)) {               ++bits;               mult *= (ll)p[j];           }       int cur = r/mult;       if (bits % 2 == 1)  sum += cur;       else  sum -= cur;    }    return r - sum;}int main(){    int T,cas = 0;    scanf("%d",&T);    while( T-- ){        int a,b,c,d,k;        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);        if(k == 0) {printf("Case %d: %d\n",++cas,0);continue;}        int t1 = b/k;        int t2 = d/k;        if( t1 > t2 ) swap(t1,t2);        ll sum = 0;        for( int i = 1 ; i <= t1 ; ++i ) sum += (ll)solve(i,t1);        sum++; sum /= 2;        for( int i = t1+1 ; i <= t2 ; ++i ) sum += (ll)solve(i,t1);        printf("Case %d: %lld\n",++cas,sum);    }    return 0;}