hdu5451_求循环节加矩阵快速幂

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5451

题目描述：

　　对于，给出x和mod，求y向下取整后取余mod的值为多少？

找循环节解析链接：http://blog.csdn.net/ACdreamers/article/details/25616461

裸题链接：http://blog.csdn.net/chenzhenyu123456/article/details/48529039

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <ctime>
#include <queue>
#include <list>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;

LL mod;
struct Mat
{
    LL a[3][3];
};
LL pow(LL base, LL n, LL mod)
{
    LL res = 1;
    while(n)
    {
        if(n % 2)
            res = (res * base) % mod;
        base = (base * base) % mod;
        n >>= 1;
    }
    return res;
}
Mat mul(Mat x, Mat y)
{
    Mat z;
    memset(z.a, 0, sizeof(z.a));
    for(int i = 0; i < 2; i++)
    {
        for(int k = 0; k < 2; k++)
        {
            for(int j = 0; j < 2; j++)
            {
                z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % mod) % mod;
            }
        }
    }
    return z;
}
void solve(LL n)
{
    Mat res, base;
    memset(res.a, 0, sizeof(res.a));
    res.a[0][0] = 1, res.a[1][1] = 1;
    base.a[0][0] = 5, base.a[0][1] = 2;
    base.a[1][0] = 12, base.a[1][1] = 5;
    while(n)
    {
        if(n % 2)
            res = mul(res, base);
        base = mul(base, base);
        n >>= 1;
    }
    LL ans = 0;
    ans = (ans + (res.a[0][0] * 5) % mod) % mod;
    ans = (ans + (res.a[1][0] * 2) % mod) % mod;
    ans = (ans * 2 - 1) % mod;
    printf("%lld
", ans);
}

int main()
{
    int t;
    LL x;
    scanf("%d", &t);
    for(int i = 1; i <= t; i++)
    {
        scanf("%lld %lld", &x, &mod);
        LL y = (mod + 1) * (mod - 1);
        LL N = pow(2, x, y);
        printf("Case #%d: ", i);
        solve(N);
    }
    return 0;
}