HDU 5451

题目要求：y = (5 + 2√6)1+2^x

设An=(5 + 2√6)n，Bn=(5 - 2√6)n

Cn = An + Bn

显然Cn是正整数，且Bn是小于1的

所以我们所求的答案其实就是Cn - 1

通过推导：

Cn * [(5 + 2√6) + (5 - 2√6)] =

((5 + 2√6)n + (5 - 2√6)n) * [(5 + 2√6) + (5 - 2√6)] =

(5 + 2√6)n + 1 + (5 - 2√6)n + 1) + (5 + 2√6)n - 1 + (5 - 2√6)n - 1)

10 * Cn = Cn + 1 + Cn - 1

Cn + 1 = 10 * Cn - Cn-1

我们找出了递推式，然后我们发现指数特别的大，M是个素数

于是我们可以实力地打一个表

/***********************************
*                                  *
*    Auther Rhapsody               *
*    E-mail addf400@foxmail.com    *
*                                  *
***********************************/
#include <set>
#include <map>
#include <cmath>
#include <deque>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

#define MP make_pair
#define PB push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 5e6 + 7;
const int INF = 0x3f3f3f3f;

int MOD = 1e9 + 7;

int f
;

//map <PII, int> Mapper;

int pow_mod(int x, int y, int MOD) {
int Ans = 1;
while (y) {
if (y & 1)
Ans = (LL)Ans * x % MOD;
x = (LL)x * x % MOD;
y >>= 1;
}
return Ans;
}

int find(int a, int idx, int MOD) {
int x = pow_mod(2, a, MOD);
x = (x + 1 - idx) % MOD;
x = (x + MOD) % MOD;
return idx - MOD + x;
}

int main(void) {
int T, Test = 0;
scanf("%d", &T);
int a, m;
while (T--) {
scanf("%d%d", &a, &m);
MOD = m;
//Mapper.clear();
f[0] = 2 % MOD;
f[1] = 10 % MOD;
//Mapper[MP(f[1], f[0])] = 1;
int idx = 2, loop, S;
while (true) {
f[idx] = (f[idx - 1] * 10 - f[idx - 2]) % MOD;
f[idx] = (f[idx] + MOD) % MOD;
//PII t = MP(f[idx], f[idx - 1]);
if (f[idx] != f[1] || f[idx - 1] != f[0]) {
//Mapper[t] = idx++;
++idx;
continue;
}
S = idx;
loop = idx - 1;
break;
}
int sum = 1;
bool flag = true;
for (int i = 1; i <= a; ++i) {
sum = sum * 2;
if (sum + 1 > S) {
flag = false;
break;
}
}
if (flag) {
printf("Case #%d: %d\n", ++Test, (f[sum + 1] - 1 + MOD) % MOD);
continue;
}
printf("Case #%d: %d\n", ++Test, (f[find(a, S, loop)] - 1 + MOD) % MOD);
}
return 0;
}