HDU 5478 Can you find it

最新推荐文章于 2019-05-05 12:09:59 发布

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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

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因为式子对所有的n都满足，我们把n=1和n=2代入.可以推出：a^k1=b^k2。b=c-a^(k1+b1) 。

我们从1到c-1枚举a。再求出b，然后判断a^k1是否等于b^k2即可。

#include<iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
using namespace std;
typedef long long ll;
ll fast(ll a,ll b,ll mod)
{
    ll res=1;
    while(b)
    {
        if(b&1)
            res=(res*a)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return res%mod;
}
int main()
{
    ll c,k1,b1,k2;
    int Case=0;
    while(cin>>c>>k1>>b1>>k2)
    {
        cout<<"Case #"<<++Case<<":"<<endl;
        ll i;
        int flag=0;
        for(i=1; i<c; i++)
        {
            ll ans=fast(i,k1,c);
            ll ans1=c-fast(i,k1+b1,c);
            ll ans2=fast(ans1,k2,c);
            if(ans==ans2)
            {
                flag=1;
                cout<<i<<" "<<ans1<<endl;
            }
        }
        if(!flag)
            puts("-1");
    }
    return 0;
}