> 0 12312
> 12312 423
> 34547 5474571
> 12312 34534
> 5645765 5675687
> 97561 123127
> 33345 34543
> 345 1
> 1 2
> 999 50000000
程序
#include<iostream>   
using namespace std;
#define ll long long

ll mod(ll a,ll b)
{
	ll t=1;
	while(b)
	{
		if(b&1==1)
			t=t*a%9901;
		a=a*a%9901;
		b>>=1;
	}
	return t;
}

ll EX_GCD(ll a,ll b,ll &x,ll &y)
{
	ll t;
	if(b==0)
	{
		x=1,y=0;
		return a;
	}
	t=EX_GCD(b,a%b,y,x);
	y-=(a/b)*x;
	return t;
}

///求求模线性方程ax=b(mod n)/////
ll modexp(ll a,ll b)
{
	ll x,y,d;
	d=EX_GCD(a,9901,x,y);
	return (x*(b/d)+9901)%9901;
}

int main()
{
	bool prime[7000+100]={false};
	ll p[1000],i,j,sum,ans_a,ans_b,a,b;
	for(i=2;i<=84;i++)
		if(prime[i]==false)
			for(j=i*i;j<=7071;j+=i)
				prime[j]=true;
	for(i=2,j=0;i<=7071;i++)
		if(prime[i]==false)
			p[j++]=i;
		p[j]=500000000+1;
	while(scanf("%lld%lld",&a,&b)!=EOF)
	{
		ans_a=ans_b=1;
		if(a==0)
		{
			cout<<"0"<<endl;
			continue;
		}
		for(i=0,j=0;p[i]<=a;i++)
		{
			if(a%p[i]==0)
			{
				sum=0;
				while(a%p[i]==0)
				{
					sum++;
					a/=p[i];
				}
				ans_a=ans_a*(mod(p[i],sum*b+1)+9900)%9901;
				ans_b=ans_b*(p[i]-1)%9901;
			}
		}
		if(a!=1)
			ans_a=ans_a*(mod(a,b+1)+9900)%9901,ans_b=ans_b*(a-1)%9901;
		printf("%lld\n",(modexp(ans_b,ans_a)+9901)%9901);				
	}
	return 0;
}

一直WA的说~~~