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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, ...).
Input
There are multiple test cases (no more than 30). For each test, a single line contains four integers C, k1, b1, k2.
Output
First, please output "Case #k: ", k is the number of test case. See sample output for more detail.
Please output all pairs (a, b) in lexicographical order. (1≤a,b<C). If there is not a pair (a, b), please output -1.
Sample Input
23 1 1 2
Sample Output
Case #1:
1 22
Source
2015 ACM/ICPC Asia Regional Shanghai Online思路：
由于(ak1*n+b1+bk2(n-1)+1)(modC)=0对于任意n为正整数恒成立，那么对于n=1成立可得(ak1+b1+b)(modC)=0;n =2时可得（a2*k1+b1+bk2+1)(modC)=0;
那么将n=1时所得的等式*ak1（modC）得（a2*k1+b1+b*ak1）（modC）=0;
那么和n=2时所得的式子比较可得（ak1）（modC）=(bk2)(modC);
那么由于1<=a,b<C;
那么从1循环到C枚举a,用快速幂求ak1，ak1+b1，用n=1时的等式求b,快速幂求bk2 ，判断是否（ak1）（modC）=(bk2)(modC);
下面证明；当a,b符合1,2式时，就(ak1*n+b1+bk2(n-1)+1)(modC)对于任意n为正整数恒成立。
1式可解得b=(C-(ak1+b1)modC);由1,2式得（ak1）（modC）=(bk2)(modC);
那么原式可改写为：(ak1*n+b1+ak1(n-1)*(C-(ak1+b1)modC))(modC)=0;
==(ak1*n+b1+ak1(n-1)*(C-(ak1+b1))(modC)=0
==(C*ak1(n-1))(modC)=0;
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
const long long int MOD=1e6+3;
#define INF 0x3f3f3f3f
#define maxn 1010
using namespace std;
#define ll long long int
ll quick(ll x,ll y);
ll C;
using namespace std;
ll quick(ll x,ll y)//快速幂
{
ll i,j,k;
i=1;
k=x;
while(y)
{
if(y&1)
{
i=(i*k)%C;
}
k=(k*k)%C;
y/=2;
}
return i;
}
int main()
{
ll i,j,k,p,q;
int e=0;
while(scanf("%lld %lld %lld %lld",&C,&k,&p,&q)!=EOF)
{
int flag=0;
e++;
printf("Case #%d:\n",e);
for(i=1;i<C;i++)
{
ll kk=quick(i,k);//a^k1,
ll k2=quick(i,k+p);//a^(k1+b1)
ll b=(C-k2);//b
ll z=quick(b,q);//b^k2;
if(z==kk)
{
flag=1;
printf("%lld %lld\n",i,b);
}
}
if(flag==0)
{
printf("-1\n");
}
}
return 0;
}