hdu 1695 容斥原理或莫比乌斯反演

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5310    Accepted Submission(s): 1907

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

 

题目大意：求在1<=x<=b,1<=y<=d上gcd(x,y)=k的(x,y)对数。

/************容斥原理*************/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

typedef __int64 LL;
const int maxn=1e5+5;
int phi[maxn],prime[maxn],factor[50],num;
bool flag[maxn];
void swap(int &a,int &b){ int t=a;a=b;b=t;}

void init()//欧拉筛选
{
    memset(flag,true,sizeof(flag));
    phi[1]=1;
    for(int i=2;i<maxn;i++)
    {
        if(flag[i])
        {
            prime[num++]=i;
            phi[i]=i-1;
        }
        for(int j=0;j<num&&i*prime[j]<maxn;j++)
        {
            flag[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);
        }
    }
}

void getfactor(int n,int &len)//质因数分解
{
    int t=sqrt(n*1.0);len=0;
    for(int i=0;i<num&&prime[i]<=t;i++)
    {
        if(n%prime[i]==0)
        {
            factor[len++]=prime[i];
            while(n%prime[i]==0) n/=prime[i];
        }
    }
    if(n>1) factor[len++]=n;
}

int getans(int a,int b)
{
    int n;
    int ans=0;
    getfactor(b,n);
    for(int i=1;i<(1<<n);i++)//容斥原理
    {
        int cnt=0,temp=1;
        for(int j=0;j<n;j++)
        {
            if(i&(1<<j))
            {
                cnt++;temp*=factor[j];
            }
        }
        if(cnt&1) ans+=a/temp;
        else ans-=a/temp;
    }
    return a-ans;
}

int main()
{
    int i,a,b,c,d,k,t,icase=0;
    LL ans;num=0;
    init();    
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d %d %d",&a,&b,&c,&d,&k);
        if(k==0||k>b||k>d)
        {
            printf("Case %d: 0
",++icase);
            continue;
        }
        ans=0;
        b/=k;d/=k;
        if(b>d) swap(b,d);
        for(i=1;i<=b;i++) ans+=phi[i];
        for(i=b+1;i<=d;i++) ans+=getans(b,i);
        printf("Case %d: %I64d
",++icase,ans);
    }
    return 0;
}
/*************莫比乌斯反演****************/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef __int64 LL;
const int maxn=1e5+5;
int prime[maxn],mu[maxn],num;
bool flag[maxn];

void init()
{
    memset(flag,true,sizeof(flag));
    mu[1]=1;
    for(int i=2;i<maxn;i++)
    {
        if(flag[i])
        {
            prime[num++]=i;mu[i]=-1;
        }
        for(int j=0;j<num&&i*prime[j]<maxn;j++)
        {
            flag[i*prime[j]]=false;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            else mu[i*prime[j]]=-mu[i];
        }
    }
}

int main()
{
    num=0;
    init();
    int i,a,b,c,d,k,t,icase=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d %d %d",&a,&b,&c,&d,&k);
        if(k==0||k>b||k>d)
        {
            printf("Case %d: 0
",++icase);
            continue;
        }
        b=b/k;d=d/k;
        if(b>d) swap(b,d);
        LL ans=0,ans1=0;
        for(i=1;i<=b;i++)
            ans+=(LL)mu[i]*(b/i)*(d/i);
        for(i=1;i<=b;i++)
            ans1+=(LL)mu[i]*(b/i)*(b/i);
        ans-=ans1/2;
        printf("Case %d: %I64d
",++icase,ans);
    }
    return 0;
}