[hoj 2255]Not Fibonacci[矩阵快速幂]

#include <stdio.h>
#include <cstring>
#include <iostream>
#define mod 10000000
using namespace std;
typedef long long LL;
LL a[4][4],tmp[4][4],ans[4][4];
//0.00 s        1260 K
/**
构造矩阵                         列向量                              初始向量
a[4][4] = { { 0,0,0,0 },
            { 0,p,q,0 },        f(n)                               f(1) = bb
            { 0,1,0,0 },        f(n-1)                             f(0) = aa
            { 0,1,0,1 } };      S(n-1)(表示从f(0)到f(n-1)的和)       S(0) = f(0) = aa
**/

void mul(LL a[][4],LL b[][4])
{
    for(int i=1; i<=3; i++)
        for(int j=1; j<=3; j++)
        {
            tmp[i][j]=0;
            for(int k=1; k<=3; k++)
            {
                tmp[i][j]+=a[i][k]*b[k][j];
                tmp[i][j]%=mod;
            }
        }
    memcpy(a,tmp,sizeof(tmp));
}
void pow(LL ans[][4],LL b[][4],int n)///矩阵快速幂,返回到ans
{
    memset(ans,0,sizeof(ans));
    for(int i=1; i<=3; i++)
        ans[i][i]=1;
    while(n)
    {
        if(n&1) mul(ans,b);
        mul(b,b);
        n>>=1;
    }
}
int main()
{
    int t,aa,bb,p,q,s,e,a1,a2;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d%d%d",&aa,&bb,&p,&q,&s,&e);
        memset(a,0,sizeof(a));
        a[1][1]=p;
        a[1][2]=q;
        a[2][1]=a[3][1]=a[3][3]=1;
        s--;
        if(s==0) a1=aa;
        else if(s<0) a1=0;
        else
        {
            memset(ans,0,sizeof(ans));
            pow(ans,a,s);
            a1=(bb*ans[3][1]+aa*ans[3][2]+aa*ans[3][3])%mod;
        }
        if(e==0) a2=aa;
        else
        {
            memset(ans,0,sizeof(ans));
            memset(a,0,sizeof(a));
            a[1][1]=p;
            a[1][2]=q;
            a[2][1]=a[3][1]=a[3][3]=1;
            pow(ans,a,e);
            a2=(bb*ans[3][1]+aa*ans[3][2]+aa*ans[3][3])%mod;
        }
        printf("%d\n",(a2-a1+mod)%mod);
    }
    return 0;
}

以上代码来源:http://blog.csdn.net/ehi11/article/details/8606751  linyiming学长

题意:

定义头两项为aa,bb的数列,

f[n] = p * f[n-1] + q * f[n-2];

求此数列第s项到第e项间的和.

思路:矩阵快速幂,构造列向量表达递推式,并且可以直接求和!

自己敲一遍:

定义矩阵结构体,重载运算符.

快速幂递归实现.

#include <cstring>
#include <cstdio>
using namespace std;
//0.00 s 	728 K
const int mod = 1e7;
typedef long long ll;
int aa,bb,p,q;
typedef struct Matrix
{
    ll A[4][4];
    Matrix()
    {
        memset(this,0,sizeof(*this));
    }
} Matrix;
ll mymod(ll x)
{
    return (x%mod + mod) %mod;
}
Matrix operator*(Matrix m1, Matrix m2)
{
    Matrix ret;
    for(int i=1; i<4; i++)
        for(int j=1; j<4; j++)
        {
            ret.A[i][j] = 0;
            for(int k=1; k<4; k++)
            {
                ret.A[i][j] += m1.A[i][k]*m2.A[k][j];
                ret.A[i][j] = mymod(ret.A[i][j]);
            }
        }
    return ret;
}
Matrix mypow(Matrix m, int n)//recursion
{
    Matrix ret, tmp;
    if(!n)
    {
        for(int i=1; i<4; i++)
            ret.A[i][i] = 1%mod;
        return ret;
    }
    tmp = mypow(m, n/2);
    if(n & 1)
        return tmp * tmp * m;
    return tmp * tmp;
}
ll sum(int n)
{
    ll ans;
    if(!n)
        ans = aa;
    else if(n==-1)
        ans = 0;
    else
    {
        Matrix m;
        m.A[1][1] = p;
        m.A[1][2] = q;
        m.A[3][1] = m.A[3][3] = m.A[2][1] = 1;
        m = mypow(m,n);
        ans = mymod(bb*m.A[3][1] + aa*m.A[3][2] + aa*m.A[3][3]);
    }
    return ans;
}
/**v(n+1) = a^n * v(1)
    取v(n+1)的第3项即S(n)
**/
int main()
{
    int t,s,e;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d %d %d %d",&aa,&bb,&p,&q,&s,&e);
        ll as,ae;
        s--;
        as = sum(s);
        ae = sum(e);
        printf("%d\n",(int)mymod(ae-as));
    }
}

快速幂循环实现.

#include <cstring>
#include <cstdio>
using namespace std;
//0.00 s 	728 K
const int mod = 1e7;
typedef long long ll;
int aa,bb,p,q;
typedef struct Matrix
{
    ll A[4][4];
    Matrix()
    {
        memset(this,0,sizeof(*this));
    }
} Matrix;
ll mymod(ll x)
{
    return (x%mod + mod) %mod;
}
Matrix operator*(Matrix m1, Matrix m2)
{
    Matrix ret;
    for(int i=1; i<4; i++)
        for(int j=1; j<4; j++)
        {
            ret.A[i][j] = 0;
            for(int k=1; k<4; k++)
            {
                ret.A[i][j] += m1.A[i][k]*m2.A[k][j];
                ret.A[i][j] = mymod(ret.A[i][j]);
            }
        }
    return ret;
}
Matrix mypow(Matrix m, int n)//loop
{
    Matrix ret;
    for(int i=1;i<4;i++)
        ret.A[i][i] = 1;
    while(n)
    {
        if(n%2) ret = ret * m;
        m = m*m;
        n /= 2;
    }
    return ret;
}
ll sum(int n)
{
    ll ans;
    if(!n)
        ans = aa;
    else if(n==-1)
        ans = 0;
    else
    {
        Matrix m;
        m.A[1][1] = p;
        m.A[1][2] = q;
        m.A[3][1] = m.A[3][3] = m.A[2][1] = 1;
        m = mypow(m,n);
        ans = mymod(bb*m.A[3][1] + aa*m.A[3][2] + aa*m.A[3][3]);
    }
    return ans;
}
/**v(n+1) = a^n * v(1)
    取v(n+1)的第3项即S(n)
**/
int main()
{
    int t,s,e;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d %d %d %d",&aa,&bb,&p,&q,&s,&e);
        ll as,ae;
        s--;
        as = sum(s);
        ae = sum(e);
        printf("%d\n",(int)mymod(ae-as));
    }
}