hdu5478 Can you find it

最新推荐文章于 2019-05-05 12:09:59 发布

原创最新推荐文章于 2019-05-05 12:09:59 发布 · 1.4k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#数论 #快速幂

各OJ刷题专栏同时被 2 个专栏收录

105 篇文章

订阅专栏

ACM—数论

27 篇文章

订阅专栏

本文探讨了给定特定模数条件下寻找满足特定方程的整数对(a, b)的问题。通过分析方程的形式与限制条件，采用快速幂算法进行有效计算。提供了一个C++实现示例，用于演示如何解决此类数学问题。

Can you find it

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 822 Accepted Submission(s): 365

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

题意：求出最小的a和对应的b满足题目所给式子。

分析：刚看到这题时也想到了快速幂，但一看数据范围，感觉肯定会超时，想了很久，未果，参考大牛博客，就是快速幂。详解见这--某大牛博客

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

ll C,k1,b1,k2;
ll a,b;

ll mod_pow(ll x, ll y, ll mod)
{
    ll re=1;
    while (y)
    {
        if (y&1) re=re*x%C;
        x=x*x%C;
        y>>=1;
    }
    return re;
}

int main ()
{
    int cas=1;
    while (cin>>C>>k1>>b1>>k2)
    {
        cout<<"Case #"<<cas++<<":"<<endl;
        bool flag=false;
        for (a=1; a<C; a++)
        {
            b = C-mod_pow(a, k1+b1, C);
            if ((mod_pow(a,k1,C)*(C-b)%C+mod_pow(b,k2,C)*b%C)%C==0)
                cout<<a<<" "<<b<<endl,flag=true;
        }
        if (!flag)
            cout<<"-1"<<endl;
    }
    return 0;
}