HDU 1695 GCD

Posted 2020-10-09 勿忘初心0924

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12935    Accepted Submission(s): 4905

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

 

 

Recommend

 

/*
* @Author: LyuC
* @Date:   2017-10-06 17:08:18
* @Last Modified by:   LyuC
* @Last Modified time: 2017-10-07 15:16:16
*/
/*
 题意：给你两个区间[a,b],[c,d],让你找有多少对(x,y)满足，x在[a,b],
    y在[c,d]，gcd(x,y)==k

 思路：因为gcd(x,y)==k,所以x/k,y/k的最大公因子是1,也就是互质，因为
    他俩最大的因子都去掉了，所以问题转化为在[1,b/k],[1,d/k]区间内，
    找有多少对互质的数
    用上莫比乌斯反演：
        设F[n]为gcd(x,y)==n的倍数的数对的数量
          f[n]为gcd(x,y)==n的数对的数量
        很显然
            F[n]=Sigma(d|n)( f(d) );
        那么
            f[n]=sigma(n|d)( u(d/n)*F[d] );

    我们要的是f[1],所以
        f[1]=u[1]*F[1]+u[2]*F[2]+...+u[min(d/k,b/k)]*F[min(d/k,b/k)];
    这种情况下多算了很多:
        假设b<d
        那么实际上是算了两边，但是不能单纯的除2，因为(x,x)这样的算了一边
        所以
            res1=从1到min(b,d)的结果
            res2=从1到max(b,d)的结果
        res=res2-res1/2;
　　　　 这个在纸上画一下就很清楚了
*/
#include <bits/stdc++.h>

#define MAXN 100005
#define LL long long

using namespace std;

int t;
int a,b,c,d;
int k;
int f[MAXN];
bool check[MAXN];
int mu[MAXN];
int prime[MAXN];
LL res1;
LL res2;

inline void mobi(){
     memset(check,false,sizeof check);
     mu[1]=1;
     int tol=0;
     for(int i=2;i<MAXN;i++){
         if(!check[i]){
             prime[tol++]=i;
             mu[i]=-1;
         }
         for(int j=0;j<tol;j++){
             if(i*prime[j]>MAXN) break;
             check[i*prime[j]]=true;
             if(i%prime[j]==0){
                 mu[i*prime[j]]=0;
                 break;
             }else{
                 mu[i*prime[j]]=-mu[i];
             }
         }
     }
}

inline void init(){
    res1=0;
    res2=0;
}

int main(){
    // freopen("in.txt","r",stdin);
    mobi();
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        init();
        printf("Case %d: ",ca);
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(k==0){
            puts("0");
            continue;
        }
        b/=k;
        d/=k;
        if(b>d){
            swap(b,d);
        }
        for(int i=1;i<=b;i++){
            res1+=(LL)mu[i]*(b/i)*(b/i);
        }
        for(int i=1;i<=b;i++){
            res2+=(LL)mu[i]*(b/i)*(d/i);
        }
        printf("%lld\n",res2-res1/2);
    }
    return 0;
}