hdoj 5478 Can you find it

Problem Description

Given a prime number C(1≤C≤2×105),
and three integers k1, b1, k2 (1≤k1,k2,b1≤109).
Please find all pairs (a, b) which satisfied the equation ak1⋅n+b1+ bk2⋅n−k2+1 =
0 (mod C)(n = 1, 2, 3, ...).

枚举a，把n=1代入，算出b。根据a，b，代入n=2，如果等式成立，就输出。

#include <iostream>     
using namespace std;     

#define ll long long

const int maxn = 100010;

ll c,k1,B1,k2;

ll p;

ll Quick_Pow(ll a,ll n) {
    ll ret=1;
    ll temp=a%p;
    while (n){
        if (n&1) ret=ret*temp%p;
        temp=temp*temp%p;
        n>>=1;
    }
    return ret;
}

int main(){
    int cas = 0;
    while(cin>>c>>k1>>B1>>k2){
        cas++;
        printf("Case #%d:\n",cas);
        bool hassol=0;
        for(int a=1;a<c;a++){
            //a^(k1*n+b1)
            //b^(k2*n-k2+1)
            p=c;
            ll a1 = Quick_Pow(a,k1+B1);
            ll b = c - a1;
            
            bool ok=1;
            ll an=a1;
            ll bn=b;
            ll atmp = Quick_Pow(a,k1);
            ll btmp = Quick_Pow(b,k2);
            an*=atmp;	bn*=btmp;
            an%=c;		bn%=c;
            if( (an+bn)%c==0 ){
                hassol=1;
                printf("%I64d %I64d\n",a,b);
            }
        }
        if(!hassol){
            printf("-1\n");
        }
    }
    return 0;
}