Can you find it

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5478

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

HINT

题意

问你有多少对数，满足a^(k1⋅n+b1) + b^(k2⋅n−k2+1) = 0 (mod C)

题解：

首先你要知道，对于每个a只有唯一对应的b可以满足这个式子，因为当n=1的时候，a^(k1+b1)+b = kk*C

由于b是小于c的，所以只有一个

所以我们可以求出b来，然后我们怎么check这个b究竟是不是呢？

随机化10个数，然后随便check就好了

代码：

//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100006
#define mod 1000000007
#define eps 1e-9
#define e exp(1.0)
#define PI acos(-1)
const double EP  = 1E-10 ;
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//*************************************************************************************

ll fMul(int m,ll n,int k)
{
    ll cc = m;
    ll b = 1;
    while (n > 0)
    {
          if (n & 1LL)
            {
                b = (b*cc);
                if(b>=k)
                    b%=k;
            }
          n = n >> 1LL ;
          cc = (cc*cc)%k;
          if(cc>=k)cc%=k;
    }
    return b;
}
int main()
{
    //freopen("out.txt","r",stdin);
    //freopen("out2.txt","w",stdout);
    srand(time(NULL));
    int tot = 1;
    int c ,k1 ,b1 ,k2;
    while(scanf("%d%d%d%d",&c,&k1,&b1,&k2)!=EOF)
    {
        printf("Case #%d:\n",tot++);
        int flag1 = 0;
        for(int i=1;i<c;i++)
        {
            int j=c-fMul(i,k1*1+b1,c);
            int flag = 1;
            for(int k=1;k<=15;k++)
            {
                ll tt = rand()%c+1;
                ll ttt1 = k1, ttt2 = k2,ttt3 = b1;
                if((fMul(i,ttt1*tt+ttt3,c)+fMul(j,ttt2*tt-ttt2+1LL,c))%c!=0)
                {
                    flag = 0;
                    break;
                }
            }
            if(flag)
            {
                printf("%d %d\n",i,j);
                flag1=1;
            }
        }
        if(!flag1)
            printf("-1\n");
    }
}