2017 Multi-University Training Contest

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题意：给你个递推式，给你N,M求F（M，1）。

一开始都没有一个递推的方向，就没有推到点上。后来看题解才有推的方向，A = {{0，1}，{2，1}}，B0 = {{1，0}，{0，1}}，B1 = {{-1，1}，{2，0}}；

官方题解：对于任意i>=1，当j>=3时，有通过归纳法可以得到

进而推导出通过矩阵快速幂求解

#include <stdio.h>#include <stdlib.h>#include <cmath>#include <string.h>#include <string>#include <algorithm>#include <map>#include <stack>#include <queue>#include <vector>#include <iostream>#define LL long long#define INF 0x3f3f3f3fconst int MAX_N = 5e2+10;const LL mod = 1e9+7;const double eps = 1e-6;using namespace std;struct Matrix{    LL a[2][2];    Matrix operator * (const Matrix &e)const    {        Matrix c;        memset(c.a,0,sizeof(c.a));        for(int i = 0;i < 2;i++)            for(int j = 0;j < 2;j++)                for(int k = 0;k < 2;k++)                    c.a[i][j] = (c.a[i][j]+a[i][k]*e.a[k][j]%mod)%mod;        return c;    }    Matrix operator - (const Matrix &e)const    {        Matrix c;        for(int i = 0;i < 2;i++)            for(int j = 0;j < 2;j++)                c.a[i][j] = (a[i][j]-e.a[i][j]+mod)%mod;        return c;    }    void MakeE()    {        for(int i = 0;i < 2;i++)            for(int j = 0;j < 2;j++)                a[i][j] = i==j?1:0;    }};Matrix Matrix_pow(Matrix ma,LL b){    Matrix tmp;    tmp.MakeE();    while(b)    {        if(b&1)            tmp = tmp*ma;        b>>=1;        ma = ma*ma;    }    return tmp;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        LL N,M;        scanf("%lld%lld",&N,&M);        Matrix A,B0,B1;        A.a[0][0] = 0; A.a[0][1] = 1;        A.a[1][0] = 2; A.a[1][1] = 1;        B0.MakeE();        B1.a[0][0] = -1; B1.a[0][1] = 1;        B1.a[1][0] = 2; B1.a[1][1] = 0;        A = Matrix_pow(A,N);        //for(int i = 0;i < 2;i++)           // for(int j = 0;j < 2;j++)               // printf(j == 1?"%d\n":"%d ",A.a[i][j]);        Matrix ma;        if(N&1)            ma = A-B1;        else            ma = A-B0;        ma = Matrix_pow(ma,M-1);        //for(int i = 0;i < 2;i++)           // for(int j = 0;j < 2;j++)               // printf(j == 1?"%d\n":"%d ",ma.a[i][j]);        LL ans = (ma.a[0][0]+ma.a[0][1]+mod)%mod;        cout<<ans<<endl;    }    return 0;}/**/

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