Coprime

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 849    Accepted Submission(s): 232


Problem Description

Please write a program to calculate the k-th positive integer that is coprime with m and n simultaneously. A is coprime with B when their greatest common divisor is 1.

 


Input

The first line contains one integer T representing the number of test cases.
For each case, there's one line containing three integers m, n and k (0 < m, n, k <= 10^9).

 


Output

For each test case, in one line print the case number and the k-th positive integer that is coprime with m and n.
Please follow the format of the sample output.

 


Sample Input

3
6 9 1
6 9 2
6 9 3

 


Sample Output

Case 1: 1
Case 2: 5
Case 3: 7
这里有两个数n,m;这里用容斥原理同样可以求在1到任意范围内，和n,m两个数互质的个数。把n,m分别分解质因数，然后把相同的合并和求一个数的本质没什么区别，套模版。然后就要去找最小的数x使得1到x与n，m互质的个数等于k，x就是答案。用二分去找n，上限直接设成1<<62#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
#include <algorithm>

using namespace std;
typedef long long int LL;
const LL INF=(LL)1<<62;
#define MAX 1000000
bool check[MAX+5];
LL prime[MAX+5];
LL sprime[MAX+5];
LL q[MAX+5];
LL n,m,k,cnt;
void eular()//线性筛
{
    memset(check,false,sizeof(check));
    int tot=0;
    for(int i=2;i<=MAX+5;i++)
    {
        if(!check[i])
            prime[tot++]=i;
        for(int j=0;j<tot;j++)
        {
            if(i*prime[j]>MAX+5) break;
            check[i*prime[j]]=true;
            if(i%prime[j]==0) break;
        }
    }
}
void Divide(LL n,LL m)
{
    cnt=0;
    LL t=(LL)sqrt(1.0*n);
    for(LL i=0;prime[i]<=t;i++)
    {
        if(n%prime[i]==0)
        {
            sprime[cnt++]=prime[i];
            while(n%prime[i]==0)
                n/=prime[i];
        }
    }
    if(n>1)
        sprime[cnt++]=n;
     t=(LL)sqrt(1.0*m);
    for(LL i=0;prime[i]<=t;i++)
    {
        if(m%prime[i]==0)
        {
            sprime[cnt++]=prime[i];
            while(m%prime[i]==0)
                m/=prime[i];
        }
    }
    if(m>1)
        sprime[cnt++]=m;
}
LL Ex(LL n)//容斥原理
{
    LL sum=0,t=1;
    q[0]=-1;
    for(LL i=0;i<cnt;i++)
    {
        LL x=t;
        for(LL j=0;j<x;j++)
        {
            q[t]=q[j]*sprime[i]*(-1);
            t++;
        }
    }
    for(LL i=1;i<t;i++)
        sum+=n/q[i];
    return sum;
}
LL Binary()
{
    LL l=1,r=INF;
    LL ans,mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if((mid-Ex(mid))>=k)
        {
            ans=mid;
            r=mid-1;
        }
        else
            l=mid+1;
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    int cas=0;
    eular();
    while(t--)
    {
        scanf("%lld%lld%lld",&m,&n,&k);
        if(n==1&&m==1)
        {
            printf("Case %d: %lld\n",++cas,k);
            continue;
        }
        Divide(n,m);
        sort(sprime,sprime+cnt);
        int cot=1;
        for(LL i=1;i<cnt;i++)
        {
            if(sprime[i]!=sprime[i-1])
            {
                sprime[cot++]=sprime[i];
            }
        }
        cnt=cot;
        printf("Case %d: %lld\n",++cas,Binary());

    }
    return 0;
}