hdu5478Can you find it

//求出所有(a^(k1*n+b1)+b^(k2*n-k2+1))%c == 0 的a,b//对于所有的n都成立//n = 1时(a^(k1+b1) + b)%c//枚举i找出对应的b//感觉应该有循环节//对于每一对成立的(a,b)找出对应的循环节//然后就过了#include<cstdio>#include<cstring>#include<iostream>#include<map>using namespace std ;const int maxn = 2e5+10 ;typedef long long ll ;ll mod ,k1 , k2 , b1 ;ll pow(ll a , ll k){    ll c = 1 ;    while(k)    {        if(k&1)c = c*a%mod ;        a = a*a%mod ;        k >>= 1 ;    }    return c ;}map<pair<ll , ll> , int>ma ;bool ans[maxn] ;int judge(ll a , ll b){    ma.clear() ;    for(ll i = 1;;i++)    {        ll t1 = pow((ll)a , k1*i+b1) ;        ll t2 = pow((ll)b , k2*i-k2+1) ;        pair<ll ,ll> tmp = make_pair(t1 , t2) ;        if((t1+t2)%mod != 0)        return -1 ;        if(ma[tmp])        return i;        ma[tmp] = 1;    }}int main(){    //freopen("in.txt" , "r" , stdin) ;    int cas = 0 ;    while(~scanf("%lld%lld%lld%lld" , &mod  , &k1 , &b1 , &k2))    {        printf("Case #%d:\n" , ++cas) ;        bool flag = false ;        for(int i = 1;i < mod;i++)        {            ll t1 = pow(i , k1+b1) ;            if(t1 == 0)continue ;            ll t2 = mod-t1 ;            if(judge(i , t2) != -1)            printf("%d %d\n" , i , t2) ,flag = true ;        }        if(!flag)puts("-1") ;    }    return 0 ;}