hdu4565 so easy 矩阵

来源：互联网发布：詹姆斯职业生涯总数据编辑：程序博客网时间：2024/05/17 17:17

的确和题目描述一样，见过这个类型就so easy了

对于这个类型，首先化为二阶通项公式的类型，然后根据特征根反过来推递推方程就行

然而无脑的我还是调了一个晚上，就因为写乘法的时候忘记return了

然后就写了两个版本的，其实并没有多大不同

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define LL long long#define fo(i,a,b) for(int i=a;i<=b;i++)#define down(i,a,b) for(int i=a;i>=b;i--)using namespace std;LL inf;#define N 2struct matrix{    LL a[N][N];    void clear()    {        fo(i,0,N-1)        fo(j,0,N-1)        a[i][j]=0;    }    matrix operator*(const matrix b)    {        matrix anss;        fo(i,0,N-1)        fo(j,0,N-1)        {            anss.a[i][j]=0;            fo(k,0,N-1)            {                anss.a[i][j]=(anss.a[i][j]+a[i][k]*b.a[k][j])%inf;                anss.a[i][j]+=inf;anss.a[i][j]%=inf;            }            anss.a[i][j]+=inf;anss.a[i][j]%=inf;        }        return anss;    }};matrix I={    1,0,    0,1};LL a,b,n;matrix KSM(matrix a,LL b){    if(b==0)return I;    if(b==1)return a;    matrix ret=KSM(a,b/2);    ret=ret*ret;    if(b%2)ret=ret*a;    return ret;}int main(){    matrix A,anss;    while(cin>>a>>b>>n>>inf)    {        if(n==1)        {            cout<<(2*a%inf)<<endl;            continue;        }        A.clear();anss.clear();        A.a[0][0]=2*a%inf;        A.a[0][1]=((b%inf-a*a%inf)%inf+inf)%inf;        A.a[1][0]=1;        A.a[1][1]=0;        anss=KSM(A,n-2);        cout<<(((anss.a[0][0]%inf*2*(a*a%inf+b%inf)%inf+2*a%inf*anss.a[0][1]%inf)%inf+inf)%inf)<<endl;    }    return 0;}

还有就是看起来精简好多的

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define LL long long#define fo(i,a,b) for(int i=a;i<=b;i++)#define down(i,a,b) for(int i=a;i>=b;i--)using namespace std;LL inf;#define N 2struct matrix{    LL a[N][N];    void clear()    {        fo(i,0,N-1)        fo(j,0,N-1)        a[i][j]=0;    }    matrix operator*(const matrix b)    {        matrix anss;        fo(i,0,N-1)        fo(j,0,N-1)        {            anss.a[i][j]=0;            fo(k,0,N-1)            {                anss.a[i][j]=((anss.a[i][j]+a[i][k]*b.a[k][j])%inf+inf)%inf;            }        }        return anss;    }};matrix I={    1,0,    0,1};LL a,b,n;matrix KSM(matrix a,LL b){    matrix ret=a,p=a;b--;    while(b)    {        if(b&1)        {            ret=ret*p;            b--;        }        p=p*p;        b>>=1;    }    return ret;}int main(){    matrix A,anss;    while(cin>>a>>b>>n>>inf)    {        if(n==1)        {            cout<<(2*a%inf)<<endl;            continue;        }        anss.clear();        A.a[0][0]=2*a;        A.a[0][1]=b-a*a;        A.a[1][0]=1;        A.a[1][1]=0;        anss=KSM(A,n-1);//        cout<<anss.a[0][0]<<' '<<anss.a[0][1]<<endl;//        cout<<anss.a[1][0]<<' '<<anss.a[1][1]<<endl;        LL ans=(anss.a[0][0]*2*a+anss.a[0][1]*2)%inf;        ans+=inf;ans%=inf;        cout<<ans<<endl;    }    return 0;}

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