hdu 5451 Best Solver 矩阵循环群+矩阵快速幂

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

题意：给定x    求解

思路： 由斐波那契数列的两种表示方法， 之后可以转化为 线性表示 F[n] = F[n-1] + F[n-2] ;

同时可以看出   和 是 一元二次方程的两根， a  = 1, b = -1 又是之后递推式的系数；

同理这里需要构造出两根为 和 ，这时 a = 1, b = –10 得 F[n] = 10F[n-1] – F[n-2]; (当然可以直接打表递推出关系式)

如果不管指数，看成是一个   这道题将变成 hdu 2256 Problem of Precision

之后需要知道如何对指数 进行取模简化，问题是具体Mod 多少？

套路是mod (M-1)*(M+1) ，具体证明详见：http://blog.csdn.net/acdreamers/article/details/25616461

到这里指数取模之后，之后跑矩阵快速幂即可；

细节: 前面矩阵快速幂原本只是跑指数-1次，正好把1抵消了；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define pb push_back
#define MK make_pair
#define A first
#define B second
#define clear0 (0xFFFFFFFE)
#define inf 0x3f3f3f3f
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define bitnum(a) __builtin_popcount(a)
#define lowbit(x) (x&(-x))
#define K(x) ((x)*(x))
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
template<typename T>
void read1(T &m)
{
    T x = 0,f = 1;char ch = getchar();
    while(ch <‘0‘ || ch >‘9‘){ if(ch == ‘-‘) f = -1;ch=getchar(); }
    while(ch >= ‘0‘ && ch <= ‘9‘){ x = x*10 + ch - ‘0‘;ch = getchar(); }
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+‘0‘);
}
inline ll gcd(ll a,ll b){ return b == 0? a: gcd(b,a%b); }
inline ll lcm(ll a,ll b){ return a/gcd(a,b)*b; }
template<class T1, class T2> inline void gmax(T1& a, T2 b){ if(a < b) a = b;}
template<class T1, class T2> inline void gmin(T1& a, T2 b){ if(a > b) a = b;}
int mod;
struct Matrix{
    int row, col;
    ll m[10][10];
    Matrix(int r,int c):row(r),col(c){ memset(m, 0, sizeof(m)); }

    bool unitMatrix(){
        if(row != col) return false;
        for(int i = 0;i < row;i++) //方阵才有单位矩阵;
                m[i][i] = 1;
        return true;
    }
    Matrix operator *(const Matrix& t){
        Matrix res(row, t.col);
        for(int i = 0; i < row; i++)
            for(int j = 0;j < t.col;j++)
                for(int k = 0; k < col; k++)
                    res.m[i][j] = (res.m[i][j] + m[i][k]*t.m[k][j])% mod;
        return res;
    }
    void print(){
        for(int i = 0;i < row; i++){
            for(int j = 0;j < col; j++)
                printf("%lld ",m[i][j]);
            puts("");
        }
    }
};

Matrix pow(Matrix a, ll n)
{
    Matrix res(a.row, a.col);
    res.unitMatrix();
    while(n){
        if(n & 1) res = res*a;
        a = a*a;
        n >>= 1;
    }
    return res;
}
ll POW(ll a,int n, ll mod)
{
    ll ans = 1;
    while(n){
        if(n&1) ans = (ans*a)%mod;
        a = a*a%mod;
        n >>= 1;
    }
    return ans;
}
int main()
{
    //freopen("data.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    Matrix mat(2,2);
    mat.m[0][0] = 5, mat.m[0][1] = 12;
    mat.m[1][0] = 2, mat.m[1][1] = 5;
    int T, kase = 1;
    scanf("%d",&T);
    while(T--){
        int n;
        read2(n, mod);
        ll MOD = 1LL*(mod-1)*(mod+1);
        MOD = POW(2,n,MOD);
        Matrix res = pow(mat, MOD);
        //res.print();
        Matrix tmp(2,1);
        tmp.m[0][0] = 5, tmp.m[1][0] = 2;
        res = res*tmp;
        printf("Case #%d: %d\n",kase++, (2*res.m[0][0]-1)% mod);
    }
    return 0;
}

ps: 这道题循环节较小，直接求解循环节也可以A；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define pb push_back
#define MK make_pair
#define A first
#define B second
#define clear0 (0xFFFFFFFE)
#define inf 0x3f3f3f3f
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define bitnum(a) __builtin_popcount(a)
#define lowbit(x) (x&(-x))
#define K(x) ((x)*(x))
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
template<typename T>
void read1(T &m)
{
    T x = 0,f = 1;char ch = getchar();
    while(ch <‘0‘ || ch >‘9‘){ if(ch == ‘-‘) f = -1;ch=getchar(); }
    while(ch >= ‘0‘ && ch <= ‘9‘){ x = x*10 + ch - ‘0‘;ch = getchar(); }
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+‘0‘);
}
inline ll gcd(ll a,ll b){ return b == 0? a: gcd(b,a%b); }
template<class T1, class T2> inline void gmax(T1& a, T2 b){ if(a < b) a = b;}
template<class T1, class T2> inline void gmin(T1& a, T2 b){ if(a > b) a = b;}

ll pow(ll a,uint n,int mod)
{
    ll ans = 1;
    while(n){
        if(n&1) ans = (ans*a)%mod;
        a = a*a%mod;
        n >>= 1;
    }
    return ans;
}
const int maxn = 463370;
int F[maxn];
int main()
{
    //freopen("data.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T, kase = 1;
    /*double d1 = 5+2*sqrt(6), d2 = 5 - 2*sqrt(6);
    rep1(i,1,10){
        printf("%.5f\n",pow(d1,i)+pow(d2,i));
    }*/
    scanf("%d",&T);
    while(T--){
        uint x, mod;
        read2(x,mod);
        F[0] = 10%mod, F[1] = 98%mod;
        ll cycle = -1;
        for(int i = 2; ; i++){
            F[i] = (10*F[i-1] - F[i-2]+ mod)% mod;
            if(F[i] == F[1] && F[i-1] == F[0]){
                cycle = i-1;
                break;
            }
        }
        int p = pow(2,x, cycle);
        printf("Case #%d: %d\n",kase++, F[p]-1);
    }
    return 0;
}