hdu 4320 Arcane Numbers 1

Arcane Numbers 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1468    Accepted Submission(s): 466

Problem Description

Vance and Shackler like playing games. One day, they are playing a game called "arcane numbers". The game is pretty simple, Vance writes down a finite decimal under base A, and then Shackler translates it under base B. If Shackler can translate it into a finite decimal, he wins, else it will be Vance’s win. Now given A and B, please help Vance to determine whether he will win or not. Note that they are playing this game using a mystery language so that A and B may be up to 10^12.

 

Input

The first line contains a single integer T, the number of test cases.
For each case, there’s a single line contains A and B.

 

Output

For each case, output “NO” if Vance will win the game. Otherwise, print “YES”. See Sample Output for more details.

 

Sample Input

3

5 5

2 3

1000 2000

 

Sample Output

Case #1: YES

Case #2: NO

Case #3: YES

 

Author

Vance and Shackler

 

Source

2012 Multi-University Training Contest 3

 

Recommend

zhoujiaqi2010

//不懂意思、到网上查了下意思、就是问A的因子数B里面是否都有

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <cmath>

#include <queue>

#include <vector>

#define N 1000000

using namespace std;

bool h[1000001];

int rc[100000];

int cnt;

void set()

{

    int i,j,k=1;

    for(i=4;i<N;i+=2)

       h[i]=1;

    for(i=3;i<1000;i+=2)

     if(!h[i])

     {

         for(j=i*i;j<N;j+=i)

           h[j]=1;

     }

    for(i=3;i<N;i+=2)

     if(!h[i]) rc[k++]=i;

      rc[0]=2;rc[k]=1000000000;

}

int main()

{

    set();

    int T,t=1;

    __int64 A,B;

    int i,k;

    __int64 c[22];//卡这里好久，A可能是超过int范围的素数、所以、悲剧

    bool b;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%I64d%I64d",&A,&B);

        int temp=sqrt(double(A));

        for(k=i=0;rc[i]<=temp;i++)

        {

             b=0;

            while(A%rc[i]==0)

            {

                b=1;

                A/=rc[i];

            }

            if(b) c[k++]=rc[i];

            if(A==1) break;

        }

        if(A!=1) c[k++]=A;

         for(i=0;i<k;i++)

          if(B%c[i])

            break;

       printf("Case #%d: ",t++);

       printf(i==k?"YES\n":"NO\n");

    }

    return 0;

}
//再插段好点的代码

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <cmath>

#include <queue>

#include <vector>

#define N 1000000

using namespace std;

bool h[1000001];

int rc[100000];

int cnt;

void set()

{

    int i,j,k=1;

    for(i=4;i<N;i+=2)

       h[i]=1;

    for(i=3;i<1000;i+=2)

     if(!h[i])

     {

         for(j=i*i;j<N;j+=i)

           h[j]=1;

     }

    for(i=3;i<N;i+=2)

     if(!h[i]) rc[k++]=i;

      rc[0]=2;rc[k]=1000000000;

}

int main()

{

    set();

    int T,t=1;

    __int64 A,B;

    int i,k;

    int c[22];

    bool b;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%I64d%I64d",&A,&B);

        int temp=sqrt(double(A));

        for(b=1,i=0;rc[i]<=temp;i++)

        {

            if(A%rc[i]==0)

            {

                if(B%rc[i]!=0)

                {

                    b=0;

                    break;

                }

                while(A%rc[i]==0)

                   A/=rc[i];

            }

         }

         if(A!=1&&(B%A)!=0) b=0;

       printf("Case #%d: ",t++);

       if(b) printf("YES\n");

           else printf("NO\n");

    }

    return 0;

}