HDU-4291 A Short problem 嵌套循环节+矩阵快速幂

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
using namespace std;

typedef long long Int64;

int MOD;

struct Matrix {
Int64 a[2][2];
inline Matrix () {
memset(a, 0, sizeof (a));
}
inline Matrix(int _a, int b, int c, int d) {
a[0][0] = _a, a[0][1] = b;
a[1][0] = c, a[1][1] = d;
}
inline Matrix operator * (Matrix y) {
Matrix ret;
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) {
for (int k = 0; k < 2; ++k) {
ret.a[i][j] += (a[i][k] * y.a[k][j]) % MOD;
}
ret.a[i][j] %= MOD;
}
}
return ret;
}
inline Matrix pow_mod(Int64 b, int mod) {
Matrix ret(1, 0, 0, 1);
MOD = mod;
while (b) {
if (b & 1) {
ret = ret * *this;
}
*this = *this * *this;
b >>= 1;
}
return ret;
}
};

// 222222224 第一个循环节，表示g[222222224]%MOD = 0, g[222222225]%MOD = 1
// 再将222222224作为MOD数，寻找下一个循环节
// 下一个循环节是 183120，原因同上
// 找到循环节后就是矩阵快速幂了

inline Int64 solve(Int64 x, int mod) {
if (x == 0) return 0;
Matrix r(3, 1, 1, 0);
return r.pow_mod(x-1, mod).a[0][0];
}

int main(  )
{
Int64 N;
while (scanf("%I64d", &N) == 1) {
printf("%I64d\n", solve(solve(solve(N, 183120), 222222224), 1000000007));
}
return 0;
}