HDU 莫比乌斯反演 1695 GCD

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

21 3 1 5 11 11014 1 14409 9

 

Sample Output

Case 1: 9Case 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).

/*hdu 1695  题意：  给出a,b,c,d,k, 求满足a <= x <= b && c <= y <= d && gcd(x,y)=k 的数对(x,y)的对数。  限制：  a=c=1; 0 < b,c <= 1e5; (n1,n2) 和 (n2,n1) 算为同种情况  思路：  其实是求满足1 <= x <= b/k && 1 <= y <= d/k && gcd(x,y)=1 的 数对(x,y)的对数。  莫比乌斯反演入门题  设f(k)为gcd(x,y)=k的数对(x,y)的对数，我们要求的是f(1)  设F(k)为gcd(x,y)为k的倍数的数对(x,y)的对数，可以想到F(k)=floor(b/k)*floor(d/k)，  由莫比乌斯反演得：  令lim=min(b/k,d/k)  f(1)=mu[1]*F(1) + mu[2]*F[2] + ... + mu[lim]*F(lim)  因为(n1,n2)和(n2,n1)算为同一种情况，所以最后结果还要减掉重复的情况。 */#include<stdio.h>#include<string.h>#include<math.h>#include <queue>#include <iostream>#include<algorithm>#define ll long longusing namespace std;const int INF=0x7fffffff;const int N=1000000+1;int prime[N],mu[N],vis[N];void Moblus(){   //O(log(n))    int i,j;    memset(vis,0,sizeof(vis));    mu[1]=1;    int cnt=0;    for (i=2;i<N;i++)    {        if (!vis[i])        {            prime[cnt++]=i;            mu[i]=-1;        }        for (j=0;j<cnt&&i*prime[j]<N;j++)        {            vis[i*prime[j]]=1;            if (i%prime[j])                mu[i*prime[j]]=-mu[i];            else{                mu[i*prime[j]]=0;                break;            }        }    }}void getMu(){  //O(nlog(n))    int i,j;    for (i=1;i<N;i++)    {        int target=i==1?1:0;        int delta=target-mu[i];        mu[i]=delta;        for (j=i+i;j<N;j+=i)            mu[i]+=delta;    }}int main(){    int i,j,a,b,c,d,k,t,p=0;    Moblus();    scanf("%d",&t);    while (t--)    {        p++;        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);        printf("Case %d: ",p);        if (k==0)        {             printf("0\n");             continue;        }        b/=k;        d/=k;        if (b>d)        {            int temp=b;            b=d;            d=temp;        }        ll ans1=0;        for (i=1;i<=b;i++)            ans1+=(ll)mu[i]*(b/i)*(d/i);        ll ans2=0;        for (i=1;i<=b;i++)      //去重            ans2+=(ll)mu[i]*(b/i)*(b/i);        ans1-=ans2/2;        printf("%lld\n",ans1);    }    return 0;}