GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17385 Accepted Submission(s): 6699

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

就是求 [1, b / k] 和 [1, d / k] 互质数的对数

先用欧拉函数求出相同区间的互质的对数多出来的用容斥去求

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;
int ans;
LL tot[maxn + ];
int prime[maxn+], phi[maxn+];
bool vis[maxn+];
void getphi()
{
    ans = ;
    phi[] = ;
    for(int i=; i<=maxn; i++)
    {
        if(!vis[i])
        {
            prime[++ans] = i;
            phi[i] = i - ;
        }
        for(int j=; j<=ans; j++)
        {
            if(i * prime[j] > maxn) break;
            vis[i * prime[j]] = ;
            if(i % prime[j] == )
            {
 
                phi[i * prime[j]] = phi[i] * prime[j]; break;
            }
            else
                phi[i * prime[j]] = phi[i] * (prime[j] - );
        }
    }
}
 
int get_cnt(int n, int m)
{
    int ans = ;
    for(int i = ; i * i <= n; i++)
    {
        if(n % i) continue;
        while(n % i == ) n /= i;
        prime[ans++] = i;
    }
    if(n != ) prime[ans++] = n;
    int res = ;
    for(int i = ; i < ( << ans); i++)
    {
        int tmp = , cnt2 = ;
        for(int j = ; j < ans; j++)
        {
            if(((i >> j) & ) == ) continue;
            tmp *= prime[j];
            cnt2++;
        }
        if(cnt2 & ) res += m / tmp;
        else res -= m / tmp;
    }
    return m - res;
}
 
int main()
{
    getphi();
    int a, b, c, d, k;
    for(int i = ; i < maxn; i++)
    {
        tot[i] = tot[i - ] + phi[i];
 
    }
    int T, kase = ;
    cin >> T;
    while(T--)
    {
        scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
        if(k == )
        {
            printf("Case %d: 0\n", ++kase);
            continue;
        }
        int n = b / k, m = d / k;
        LL sum = tot[n > m ? m : n];
       // cout << sum << endl;
        if(m > n) swap(n, m);
        for(int i =m + ; i <= n; i++)
        {
            sum += get_cnt(i, m);
        }
        printf("Case %d: %lld\n", ++kase, sum);
    }
 
    return ;
}

GCD