hdu 1695 GCD 莫比乌斯

GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9765 Accepted Submission(s): 3652


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, print the number of choices. Use the format in the example.

[align=left]Sample Input[/align]

2
1 3 1 5 1
1 11014 1 14409 9

[align=left]Sample Output[/align]

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).

[align=left]Source[/align]
2008 “Sunline Cup” National Invitational Contest

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 0.00000000001
#define pi 4*atan(1)
const int N=1e5+10,M=1e7+10,inf=1e9+10,mod=1e9+7;
int mu
, p
, np
, cnt, sum
;
void init() {
mu[1]=1;
for(int i=2; i<N; ++i) {
if(!np[i]) p[++cnt]=i, mu[i]=-1;
for(int j=1; j<=cnt && i*p[j]<N; ++j) {
int t=i*p[j];
np[t]=1;
if(i%p[j]==0) { mu[t]=0; break; }
mu[t]=-mu[i];
}
}
}
int main()
{
int T,cas=1;
init();
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: 0\n",cas++);
continue;
}
b/=k,d/=k;
if(b>=d)swap(b,d);
ll ans=0;
for(int i=1;i<=b;i++)
{
ans+=(ll)mu[i]*(b/i)*(d/i);
}
ll ans2=0;
for(int i=1;i<=b;i++)
{
ans2+=(ll)mu[i]*(b/i)*(b/i);
}
printf("Case %d: %lld\n",cas++,ans-ans2/2);
}
return 0;
}