hdu 5478 Can you find it

最新推荐文章于 2021-04-05 09:08:08 发布

原创最新推荐文章于 2021-04-05 09:08:08 发布 · 165 阅读

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本文探讨了给定特定条件下，寻找满足模数方程的所有有序数对(a, b)的方法。通过假设n=1来简化问题，并使用快速幂运算进行求解验证。

Can you find it

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1967 Accepted Submission(s): 832

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

思路：
因为对于所有n都成立，所以先假设n==1,
便有(a^(k1+b1)+b)%C==0
枚举a,得出b的解，对于每对(a,b),验证是否对所有n都成立。
将n==2带入，若成立，则所有n都成立(我猜的，不要问为什么)。

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll C, k1, k2, b1;
ll pow1(ll a, ll b, ll mod)
{
    ll r = 1;
    while (b)
    {
        if (b & 1) r = r * a % mod;
        a = a * a % mod;
        b /= 2;
    }
    return r;
}
int main()
{
    int cas = 0;
    while (~scanf("%lld%lld%lld%lld", &C, &k1, &b1, &k2))
    {
        bool bb=0;
        printf("Case #%d:\n", ++cas);
        for (ll i = 1; i < C; i++)
        {
            ll a = pow1(i, k1 + b1, C);
            ll b = C - a;
            if (b >= 1 && b < C)
            {
                if ((pow1(i, 2 * k1 + b1, C) + pow1(b, k2 + 1, C)) % C == 0)
                    printf("%lld %lld\n", i, b),bb=1;
            }
        }
        if(!bb)printf("-1\n");
    }
    return 0;
}