[HDU 1695] GCD

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7002    Accepted Submission(s): 2577

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

 

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).

 

Source

2008 “Sunline Cup” National Invitational Contest

 

容斥定理、具体见代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define ll long long
#define N 100000

int tot;
int prime[N+10];
bool isprime[N+10];
int phi[N+10];
void prime_pri()
{
    tot=0;
    phi[1]=1;
    memset(isprime,true,sizeof(isprime));
    isprime[0]=isprime[1]=false;
    for(int i=2;i<=N;i++)
    {
        if(isprime[i])
        {
            prime[tot++]=i;
            phi[i]=i-1;
        }
        for(int j=0;j<tot;j++)
        {
            if(i*prime[j]>N) break;
            isprime[i*prime[j]]=false;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else
            {
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
}
int fatcnt;
int factor[N][2];
int getfactors(int x)
{
    fatcnt=0;
    int tmp=x;
    for(int i=0;prime[i]<=tmp/prime[i];i++)
    {
        factor[fatcnt][1]=0;
        if(tmp%prime[i]==0)
        {
            factor[fatcnt][0]=prime[i];
            while(tmp%prime[i]==0)
            {
                factor[fatcnt][1]++;
                tmp/=prime[i];
            }
            fatcnt++;
        }
    }
    if(tmp!=1)
    {
        factor[fatcnt][0]=tmp;
        factor[fatcnt++][1]=1;
    }
    return fatcnt;
}
int cal(int n,int m) //求1到n中与m互质的数的个数
{
    int tmp,cnt,ans=0;
    getfactors(m);
    for(int i=1;i<(1<<fatcnt);i++) //0表示不选择因子
    {
        cnt=0;
        tmp=1;
        for(int j=0;j<fatcnt;j++)
        {
            if(i&(1<<j))
            {
                cnt++;
                tmp*=factor[j][0];
            }
        }
        if(cnt&1) ans+=n/tmp;
        else ans-=n/tmp;
    }
    return n-ans;
}
int main()
{
    prime_pri();
    int T,iCase=1;
    int a,b,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d%d",&a,&a,&b,&b,&k);
        if(k==0) //除0特判
        {
            printf("Case %d: 0
",iCase++);
            continue;
        }
        a/=k,b/=k;
        if(a>b) swap(a,b);
        ll ans=0;
        for(int i=1;i<=b;i++)
        {
            if(i<=a) ans+=phi[i];
            else ans+=cal(a,i);
        }
        printf("Case %d: %lld
",iCase++,ans);
    }
    return 0;
}