1 solutions

  • 0
    @ 2025-11-27 11:58:50

    C++ :

    #include<bits/stdc++.h>
    using namespace std;
    long long t,a,p;
    long long ys[100010];
    long long fpow(long long a,long long b,long long mod)
    {
    	if(b==0) return 1;
    	if(b==1) return a%mod;
    	if(b&1)  return a*fpow(a*a%mod,b>>1,mod)%mod;
    	else return fpow(a*a%mod,b>>1,mod)%mod;
    }
    int main()
    {
    	cin>>t;
    	while(t--)
    	{
    		cin>>a>>p;
    		if(fpow(a,p-1,p)!=1)printf("No\n");
    		else
    		{
    			int f=0,t=0;
    			//找除了1和本身以外的其它因数,存起来
    			for(int i=2;i*i<=p-1;i++)
    			{
    				if((p-1)%i==0)
    				{
    					ys[++t]=i;
    					ys[++t]=(p-1)/i;
    				}
    			}
    			for(int i=1;i<=t;i++)
    			{
    				if(fpow(a,ys[i],p)==1)
    				{
    					printf("No\n");
    					f=1;
    					break;
    				}
    			}
    			if(f==0)
    			{
    				printf("Yes\n");
    			}
    		}
    	}
    	return 0;
    }
    
    
    
    • 1

    Information

    ID
    90
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
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