GCD

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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.

You 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).

分析：gcd（x，y）=k，等价于gcd（x/k，y/k）=1；

　　　所以把区间缩小k倍，然后枚举小的数用容斥求出答案即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,cnt,fac[maxn],cas;
int x,y;
void init(int x)
{
    cnt=0;
    if(x%2==0){
        fac[++cnt]=2;
        while(x%2==0)x/=2;
    }
    for(int i=3;(ll)i*i<=x;i+=2)
    {
        if(x%i==0)
        {
            fac[++cnt]=i;
            while(x%i==0)x/=i;
        }
    }
    if(x>1)fac[++cnt]=x;
}
ll gao(ll x)
{
    ll ret=0;
    for(int i=1;i<(1<<cnt);i++)
    {
        ll num=0,now=1;
        for(int j=0;j<cnt;j++)
        {
            if(i&(1<<j))
            {
                ++num;
                now*=fac[j+1];
            }
        }
        if(num&1)ret+=x/now;
        else ret-=x/now;
    }
    return x-ret;
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d%d",&x,&x,&y,&y,&k);
        if(!k)
        {
            printf("Case %d: 0
",++cas);
            continue;
        }
        x/=k,y/=k;
        if(x>y)swap(x,y);
        ll ret=0;
        rep(i,1,x)
        {
            init(i);
            ret+=gao(y)-gao(i-1);
        }
        printf("Case %d: %lld
",++cas,ret);
    }
    return 0;
}