HDU1695(容斥原理)

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9811    Accepted Submission(s): 3682

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

 

思路：题意可转化为求[1,b/k]与[1,d/k]组成数对(x,y)。x,y互质的对数。当x与y均不大于min(b/k,d/k)时，需要将答案除以2。

#include <cstdio>
#include <algorithm> 
#include <vector>
using namespace std;
const int MAXN=100001;
typedef long long LL;
LL b,d,k;
vector<LL> divisor[MAXN];
void prep()
{
    for(LL e=1;e<MAXN;e++)
    {
        LL x=e;
        for(LL i=2;i*i<=x;i++)
        {
            if(x%i==0)
            {
                divisor[e].push_back(i);
                while(x%i==0)    x/=i;
            }
        }
        if(x>1)    divisor[e].push_back(x);
    }
}
LL sieve(LL m,LL n)
{
    LL ans=0;
    for(LL mark=1;mark<(1<<divisor[n].size());mark++)
    {
        LL mul=1;
        LL odd=0;
        for(LL i=0;i<divisor[n].size();i++)
        {
            if(mark&(1<<i))
            {
                mul*=divisor[n][i];
                odd++;
            }
        }
        LL cnt=m/mul;
        if(odd&1)    ans+=cnt;
        else    ans-=cnt;
    }
    return m-ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    prep();
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%*d%lld%*d%lld%lld",&b,&d,&k);
        printf("Case %d: ",cas);
        if(k==0)
        {
            printf("%d
",0);
            continue;
        } 
        b/=k;
        d/=k;
        if(b>d)    swap(d,b);
        LL res=0;
        for(LL i=1;i<=b;i++)
        {
            LL cnt=sieve(b,i);
            res+=cnt;
        }
        res=(res+1)/2;
        for(LL i=b+1;i<=d;i++)
        {
            LL cnt=sieve(b,i);
            res+=cnt;
        }
        printf("%lld
",res);
    }
    return 0;
}