Codeforces186C

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#矩阵快速幂

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本文介绍了一个关于三角形植物繁殖问题的算法题解，通过矩阵快速幂的方法求解经过n年后指向上的三角形植物数量，并给出了具体的C++实现代码。

题目来源：Codeforces186C

题目描述：

C. Plant

Dwarfs have planted a very interesting plant, which is a triangle directed “upwards”. This plant has an amusing feature. After one year a triangle plant directed “upwards” divides into four triangle plants: three of them will point “upwards” and one will point “downwards”. After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.

MarkdownPicture
Help the dwarfs find out how many triangle plants that point “upwards” will be in n years.

Input
The first line contains a single integer n (0 ≤ n ≤ 1018) — the number of full years when the plant grew.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output
Print a single integer — the remainder of dividing the number of plants that will point “upwards” in n years by 1000000007 (109 + 7).

Examples
input
1
output
3
input
2
output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.

time limit per test2 seconds
memory limit per test256 megabytes

题解：

矩阵快速幂，递推关系式为 an=2*an-1+ 4^(n-2),然后退出矩阵即可

题目分类：

矩阵快速幂

代码：

//zcc
#include<iostream>
#include<cstring>
#include<algorithm>
#include<stdio.h>
using namespace std;

typedef long long ll;
const  ll mod= 1000000007;

ll mod_pow(ll x,ll n)
{
    ll res=1;
    while(n>0)
    {
        if(n&1)res=res*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    return res;
}
typedef struct node{
    ll edge[3][3];
}Matrix;
ll n,m=10000;
Matrix mz,ant,h;

void Mult(Matrix &a,Matrix &b,Matrix &c)
{
    int i,j,k;
    memset(h.edge,0,sizeof(h.edge));
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        for(k=0;k<2;k++)
    {
        h.edge[i][j]+=a.edge[i][k]*b.edge[k][j];
        h.edge[i][j]%=mod;
    }
    for(i=0;i<2;i++)for(j=0;j<2;j++)c.edge[i][j]=h.edge[i][j];
}

void KSM(Matrix a,ll k)
{
    while(k>=1)
    {
        if(k&1)Mult(ant,mz,ant);
        Mult(mz,mz,mz);
        k>>=1;
    }
}

int main()
{
   cin>>n;
    {
         mz.edge[0][0]=2;
        mz.edge[0][1]=0;
        mz.edge[1][0]=4;
        mz.edge[1][1]=4;
        ant.edge[0][0]=10;
        ant.edge[0][1]=4;
        if(n==0)cout<<1<<endl;
        else if(n==1)cout<<3<<endl;
        else if(n==2)cout<<10<<endl;
        else
        {
            KSM(ant,n-2);cout<<ant.edge[0][0];
        }
    }
    return 0;
}