HDU 5478 Can you find it

Can you find it

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 450    Accepted Submission(s): 208

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

 

Recommend

hujie

 

 

首先想n为1,2,3……都要成立，所以先保证n=1成立，然后验证其他是否成立，把等式左边的b那一项移到右边，除一下，发现每次增量都是a^k1 和b^k2 所以只要n=2成立，后面的都成立。

下回这种题不要怕，还是可以做的。

 

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 1234567
#define M 12
int c, k1, b1, k2;
bool flag;
int quickpow(int m,int n,int k)// m^n % k
{
    int  b = ;
    while (n > )
    {
          if (n & )
             b = (b*m)%k;
          n = n >>  ;
          m = (m*m)%k;
    }
    return b;
}
int main()
{
    int tt = ; int b;
    while(~scanf("%d %d %d %d", &c, &k1, &b1, &k2))
    {

        printf("Case #%d:\n", tt++);
        flag = ;
        for(int a = ; a < c; a++)
        {
            b = c - quickpow(a,k1+b1,c);
            if( ( quickpow(a, k1*+b1, c) + quickpow(b, k2*-k2+, c) ) %c == )
            {
                printf("%d %d\n",a,b);
                flag = ;
            }
        }
        if(flag == )
            printf("-1\n");
    }
    return ;
}