矩阵基础1002 CodeForces 450B

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题意:
F[i]=F[i-1]+F[i+1]
求F[n]
思路:
F[i+1]=F[i]-F[i-1]->F[i]=F[i-1]-F[i-2]
然后矩阵快速幂去做
这道题有些特殊性,简单推导一下就会发现
每6次一个循环,打个小表就行了

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>#include<queue>#include<stack>#include<string>#include<vector>#include<map>#include<set>using namespace std;#define lowbit(x) (x&(-x))typedef long long LL;const int maxn = 100005;const int inf=(1<<28)-1;#define Matrix_Size 5const LL MOD = 1e9+7;int Size;struct Matrix{    LL mat[Matrix_Size][Matrix_Size];    void clear()    {        memset(mat,0,sizeof(mat));    }    void output()    {        for(int i = 0;i < Size;i++)        {            for(int j = 0;j < Size;j++)                printf("%d ",mat[i][j]);            printf("\n");        }    }    Matrix operator *(const Matrix &b)const    {        Matrix ret;        for(int i = 0;i < Size;i++)            for(int j = 0;j < Size;j++)            {                ret.mat[i][j] = 0;                for(int k = 0;k < Size;k++)                {                    long long tmp = (long long)mat[i][k]*b.mat[k][j]%MOD;                    ret.mat[i][j] = (ret.mat[i][j]+tmp);                    if(ret.mat[i][j]>=MOD)                        ret.mat[i][j] -= MOD;                    if(ret.mat[i][j]<0)//注意是否需要MOD                         ret.mat[i][j] += MOD;                }            }        return ret;    }};Matrix pow_M(Matrix a,long long n){    Matrix ret;    ret.clear();    for(int i = 0;i < Size;i++)        ret.mat[i][i] = 1;    Matrix tmp = a;    while(n)    {        if(n&1)ret = ret*tmp;        tmp = tmp*tmp;        n>>=1;    }    return ret;}int main(){    LL x,y,n;    scanf("%lld%lld%lld",&x,&y,&n);    if(n==1)    {        printf("%lld\n",(x+MOD)%MOD);        return 0;    }    Size=2;    Matrix A,B;    A.mat[0][0]=x,A.mat[0][1]=y;    A.mat[1][0]=0,A.mat[1][1]=0;    B.mat[0][0]=0,B.mat[0][1]=-1;    B.mat[1][0]=1,B.mat[1][1]=1;    B=pow_M(B,n-2);    A=A*B;    //B.output();    //printf("%lld\n",((x*B.mat[1][0])%MOD+(y*B.mat[1][1])%MOD+MOD)%MOD);    //printf("%lld\n",A.mat[0][1]);    printf("%lld\n",(A.mat[0][1]+MOD)%MOD);    return 0;}

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