hdu1695GCD

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4737    Accepted Submission(s): 1693

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

// 在(1,b),(1,d)里面各取一个数 使得gcd(x,y) == k
//即 n*k == x && m*k == y && gcd(x,y) == 1 ；
//移项 n = x/k , m = y/k ; 所以转化为在 (1,b/k),和(1,d/k)各取一个数
//使得这两个数互素有多少种取法 令 n = b/k , m = d/k , n <= m
//因为(s,e)和(e,s)是相同的，所以我们可以枚举 m ,设枚举到 y
// 在 y <= n 时,和y 互素的就是 phi[y] 了，在 y > n 时 就比较麻烦、
// 我们可以把 y 分解成 p1^x1*p2^x2*p3^x3 。。。然后求(1,n) 中和 y不互素的数
// 也就是把(1,n) 中 p1,p2,p3.。。的倍数给去掉，去掉的时候会重复，就需要用到容斥定理来求

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
#define maxn 100010
#define LL long long
LL phi[maxn] ;
vector<int>qe[maxn] ;
void init()
{
    int n , i ;
    n = maxn-10 ;
    for( i = 1 ; i <= n ;i++)phi[i] = i ;
    for( i = 2 ; i <= n ;i++)if(phi[i] == i )
    {
        for( int j = i ; j <= n ;j += i ){
          phi[j] = phi[j]*(i-1)/i;
          qe[j].push_back(i) ;
        }
    }
    for( i = 2 ; i <= n ;i++ )
        phi[i] = phi[i]+phi[i-1] ;
}
LL dfs( int x , int n , int now )
{
    int i ;
    LL ret = 0 ;
    for( i = x ; i < qe[now].size() ; i++ )
    {
        ret += n/qe[now][i]-dfs(i+1,n/qe[now][i],now) ;
    }
    return ret ;
}
int main()
{
    int i , n , m ,num ,k , T , case1 = 0 ;
    int a , b , c , d , e ;
    LL ans ;
    init() ;
   // freopen("in.txt","r",stdin) ;
    cin >> T ;
    while(T--)
    {
        cin >> a >> b >> c >> d >> k ;
        if(k == 0)
        {
            printf("Case %d: ",++case1) ;
            puts("0") ;
            continue ;
        }
        if( b > d ) swap(d,b) ;
        n = b/k ; m = d/k ;
        ans = phi[n] ;
        for( i = n+1 ; i <= m ;i++ )
        {
            if(phi[i] == i-1) ans += n ;
            else
            ans += n-dfs(0,n,i) ;
        }
        printf("Case %d: ",++case1) ;
        cout << ans << endl;
    }
    return 0 ;
}