HDU5478 Can you find it

題目鏈接

題意：給定 C，k1, b1, k2 找出所有的（a, b）滿足 ak1⋅n+b1+ bk2⋅n−k2+1 = 0 (mod C)(n = 1, 2, 3, ...)  （1<=a, b <C）

思路：n = 1時，找出（1，C）範圍內的滿足的(a, b)組合，然後暴力枚舉n，把符合的留下

代碼如下：

(1≤a,b<C)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
typedef __int64 ll;
const int N = 1e5+10;
ll C, k1, b1, k2;
ll a[2*N], b[2*N];

ll pow_mod(ll x, ll y, ll k)
{
	ll s = 1;
	while(y)
	{
		if (y % 2)
			s = (s * x)%k;
		x = (x * x)%k;
		y >>= 1;
	}
	return s;
}

int main()
{  
	int cas = 1;
	while(~scanf("%I64d%I64d%I64d%I64d", &C, &k1, &b1, &k2))
	{
		printf("Case #%d:\n", cas++);
		for(ll i = 1; i < C; i++)
		{
			ll cnt = pow_mod(i, k1 + b1, C);
			a[i] = i;
			b[i] = (cnt / C + 1) * C - cnt;
		}
		bool flag; ll ans = 0;
		for(ll i = 1; i < C; i++)
		{
			flag = 0;
			for(ll j = 2; j <= 100; j++)
			{
				if((pow_mod(a[i], k1 * j + b1, C) + pow_mod(b[i], k2 * (j-1) + 1, C)) % C)
				{    
					flag = 1;
					break;
				}
			}
			if(!flag)
			{
				ans++; printf("%I64d %I64d\n", a[i], b[i]);
			}
		}
		if(ans == 0) printf("-1\n");
	}
	return 0;	
}