HDU5478——Can you find it(化简)

Description：

Given a prime number $C(1≤C≤2×10^5)$, and three integers $k_{1}, b_{1}, k_{2} (1≤k1,k2,b1≤10^9)$. Please find all pairs $(a, b)$ which satisfied the equation $a^{k_{1}⋅n+b_{1}} + b^{k_{2}⋅n−k_{2}+1} = 0 (mod C)(n = 1, 2, 3, ...)$.

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

题意：

给你一个式子 $a^{k_{1}⋅n+b_{1}} + b^{k_{2}⋅n−k_{2}+1} = 0 (mod C)(n = 1, 2, 3, ...)$.求出这样一对整数对 $(a,b)$ 使得对任意的n使此等式成立,并将 $(a,b)$ 按字典序输出，如果没有满足要求的这样的整数对，就输出 $-1$

当 $n=1$ 时， $a^{k_{1}+b_{1}} + b = 0 (mod C)$ ①

当 $n=2$ 时， $a^{2*k_{1}+b_{1}} + b^{k_{2}+1} = 0 (mod C)$ ②

对 ① 式两边同乘 $a^{k_{1}}$得： $a^{k_{1}} *(a^{k_{1}+b1} + b) = 0 (mod C)$

即： $a^{2*k_{1}+b_{1}} + a^{k_{1}} * b = 0 (mod C)$ ③

结合 ② ③ 得 $a^{2*k_{1}+b_{1}} + a^{k_{1}} * b = a^{2*k_{1}+b_{1}} + b^{k_{2}+1} = 0 (mod C)$ ④

化简 ④ 式得 $a^{k_{1}} = b^{k_{2}}$ ⑤

对于每一个a的值用快速幂求出 $a^{k_{1}+b_{1}}$，然后由①式得到 $b$

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(int m, int k, int mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

const int N = 2000010;
int phi[N];
void oula()
{
    phi[1] = 1;
    for (int i = 2; i < N; i++)
    {
        phi[i] = i;
    }
    for (int i = 2; i < N; i++)
    {
        if (phi[i] == i)
        {
            for (int j = i; j < N; j += i)
            {
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    }
}

ll c, k1, k2, b1;
int main()
{
    int cas = 0;
    while (~scanf("%lld%lld%lld%lld", &c, &k1, &b1, &k2))
    {
        cas++;
        printf("Case #%d:\n", cas);
        bool flag = 0;
        for (ll i = 1; i < c; i++)
        {
            ll res = qpow(i, k1, c);
            ll b = c - qpow(i, k1 + b1, c);
            ll temp = qpow(b, k2, c);
            if (res == temp)
            {
                flag = 1;
                pldd(i, b);
            }
        }
        if (!flag)
            pd(-1);
    }

    return 0;
}