hdu1695 GCD2 容斥原理 求x属于[1,b]与y属于[1,d]，gcd(x,y)=k的对数。(5,7)与(7,5)看作同一对。

GCD
Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)
Total Submission(s):     Accepted Submission(s): 
 
Problem Description
Given  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=, y=) and (x=, y=) are considered to be the same.
 
Yoiu can assume that a = c =  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 , cases.
Each case contains five integers: a, b, c, d, k,  < a <= b <= ,,  < c <= d <= ,,  <= k <= ,, as described above.
 
Output
For each test case, print the number of choices. Use the format in the example.
 
Sample Input
 
Sample Output
 
Case :
Case : 
 
Hint
For the first sample input, all the  pairs of numbers are (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ), (, ).
 
/**
题目：hdu1695 GCD2
链接：http://acm.hdu.edu.cn/showproblem.php?pid=1695
题意：求x属于[1,b]与y属于[1,d]，gcd(x,y)=k的对数。(5,7)与(7,5)看作同一对。
思路：
gcd(x,y)=k => gcd(x/k,y/k) = 1;
 
则求x/k与y/k互质对数。
 
即求：[1,b/k]与[1,d/k]之间互质的对数
 
设x属于[1,b/k], y属于[1,d/k];
枚举x，求x与y互质的对数。所以要预处理所有x的质因子。然后容斥处理。由于(x,y)=>(5,7),(7,5)是同一组。
所以：答案为ans += (d/k) - x在d/k中不互质的数 - (x-1);
*/
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#define LL long long
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5+;
vector<int> prime[maxn];
int flag[maxn];
void init()
{
    memset(flag, , sizeof flag);
    for(ll i = ; i < maxn; i++){
        if(flag[i]==){
            prime[i].push_back(i);
            for(ll j = *i; j < maxn; j+=i){
                prime[j].push_back(i);
                flag[j] = ;
            }
        }
    }
}
ll rc(int pos,int n)
{
    ll sum = ;
    ll mult, ones;
    ll len = prime[pos].size();
    ll m = <<len;
    for(int i = ; i < m; i++){
        ones = ;
        mult = ;
        for(int j = ; j < len; j++){
            if(i&(<<j)){
                ones++;
                mult = mult*prime[pos][j];
                if(mult>n) break;
            }
        }
        if(ones%==){
            sum -= n/mult-(pos-)/mult;
        }else
        {
            sum += n/mult-(pos-)/mult;
        }
    }
    return n-(pos-)-sum;
}
int main()
{
    init();
    int T;
    int cas = ;
    int a, b, c, d, k;
    cin>>T;
    while(T--)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(k==){
            printf("Case %d: 0\n",cas++);continue;
        }
        if(b>d) swap(b,d);///for b<=d;
        b = b/k;
        d = d/k;
        ll ans = ;
        if(b>=){
            ans += d;
        }
        for(int i = ; i <= b; i++){
            ans += rc(i,d);
        }
        printf("Case %d: %lld\n", cas++,ans);
    }
    return ;
}