【hdu 1695】GCD（莫比乌斯反演）

GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10201 Accepted Submission(s): 3849

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

HintFor 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

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[题意][求x∈[a,b]，y∈[c,d]时，gcd(x，y)=k的x，y的个数，gcd(x，y)和gcd(y，x)算一种.在本题中，a和c恒为1]
【题解】

∑x=1b∑y=1dgcd(x,y)=k

=∑x=1b∑y=1d[k|i][d|j][(ik,jk)=1]

设 x′=kx，y′=ky，

=∑x′=1bk∑y′=1dk[(x′，y′)=1]

∵[n=1]=∑d|nμ(d)

∴设bk=n，dk=m，x=x′,y=y′，N=min(n,m)

则=∑d=1N∑x=1n∑y=1m[d|x][d|y]μ(d)

=∑d=1N∑x=1n[d|x]∑y=1m[d|y]μ(d)

=∑d=1N⌊xd⌋⌊yd⌋μ(d)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int prime[100010],miu[100010];
int a,b,c,d,k,t;
bool p[100010];
inline void shai(int n)
{
    int i,j;
    miu[1]=1;
    for (i=2;i<=n;++i)
     {
         if(!p[i]) prime[++prime[0]]=i,miu[i]=-1;
         for (j=1;j<=prime[0];++j)
          {
              if (i*prime[j]>n) break;
              p[i*prime[j]]=1;
              if (!(i%prime[j])) {miu[i*prime[j]]=0;break;}
               else miu[i*prime[j]]=-miu[i];
          }
     }
}
int main()
{
    int i,j;
    scanf("%d",&t);
    shai(100000);
    for (i=1;i<=t;++i)
     {
         scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
         if (!k) {printf("Case %d: 0\n",i); continue;}
         long long ans1=0,ans2=0;
         b/=k; d/=k;
         if (b>d) swap(b,d);
         for (j=1;j<=b;++j) ans1+=(long long)(b/j)*(d/j)*miu[j];
         for (j=1;j<=b;++j) ans2+=(long long)(b/j)*(b/j)*miu[j];
         ans1-=ans2/2;
         printf("Case %d: %I64d\n",i,ans1);
     }
    return 0;
}