HDU4382(特殊的矩阵连乘)

题目：Harry Potter and Cyber Sequence Generator

题意，有两个容器C1,C2,初始的时候C1中有一个数的值为V，给你K个操作，每次都重复这K个操作N遍，最后问你C2中的数是

     多少。N<=10^100。

1：循环操作的次数巨大，敏感的想到这是矩阵连乘的题目。

2：K个操作可以得出一个矩阵，N个K操作就是这个矩阵的N次方

3：最后再乘以初始矩阵即可

#include 
#include 
#include 

using namespace std;
typedef long long LL;

const LL MOD=1000000007;
const int N=3;

struct Matrix
{
    LL m[N][N];
};

Matrix per={1,0,0,0,1,0,0,0,1};

Matrix multi(Matrix a,Matrix b)
{
    Matrix c;
    int i,j,k;
    for(i=0;i>=1;
        p=multi(p,p);
    }
    return ans;
}

LL getNum(char *str)
{
    LL ans=0;
    int len=strlen(str);
    for(int i=0;i>T;
    while(T--)
    {
        cin>>V;
        ans=per;
        while(cin>>str1)
        {
            if(str1[0]=='E') break;
            cin>>str2>>str3;
            tmp1=per;
            if(str1[0]=='S')
            {
                if(strcmp(str2,"C1,")==0&&strcmp(str3,"C2")==0)
                {
                    tmp1.m[0][0]=0;
                    tmp1.m[1][0]=1;
                }
                else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C1")==0)
                {
                    tmp1.m[0][1]=1;
                    tmp1.m[1][1]=0;
                }
                else if(strcmp(str2,"C1,")==0&&strcmp(str3,"C1")!=0)
                {
                    tmp1.m[0][0]=0;
                    tmp1.m[2][0]=getNum(str3);
                }
                else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C2")!=0)
                {
                    tmp1.m[1][1]=0;
                    tmp1.m[2][1]=getNum(str3);
                }
            }
            if(str1[0]=='A')
            {
                if(strcmp(str2,"C1,")==0&&strcmp(str3,"C1")==0)
                    tmp1.m[0][0]=2;
                else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C2")==0)
                    tmp1.m[1][1]=2;
                else if(strcmp(str2,"C1,")==0&&strcmp(str3,"C2")==0)
                    tmp1.m[1][0]=1;
                else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C1")==0)
                    tmp1.m[0][1]=1;
                else if(strcmp(str2,"C1,")==0)
                    tmp1.m[2][0]=getNum(str3);
                else
                    tmp1.m[2][1]=getNum(str3);
            }
            if(str1[0]=='M')
            {
                if(strcmp(str2,"C1,")==0)
                    tmp1.m[0][0]=getNum(str3);
                else
                    tmp1.m[1][1]=getNum(str3);
            }
            ans=multi(ans,tmp1);
        }
        cin>>Q;
        cout<<"Case "<>str;
            LL ret=0;
            int len=strlen(str);
            for(int i=0;i