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;
}