HDU 2604

来源：互联网发布：c语言面向对象编程 pdf 编辑：程序博客网时间：2024/06/04 19:51

快速幂取模，快速矩阵幂取模

先利用DFA构造转移矩阵

详见AC博客http://hi.baidu.com/aekdycoin/item/f3a474a7ee3b0d218919d3ae

#include<cstdio>#include<cstring>using namespace std;typedef long long ll;int len,mod;ll tem[6][6];int main(){    int i,j,k;    while(scanf("%d %d",&len,&mod)==2){        ll a[6][6]={{0,0,1,0,0,1},{0,0,0,0,0,1},{0,0,0,0,1,0},{0,0,0,1,1,0},{0,2,0,1,0,0},{2,0,1,0,0,0}};        ll t[6][6]={{1,0,0,0,0,0},{0,1,0,0,0,0},{0,0,1,0,0,0},{0,0,0,1,0,0},{0,0,0,0,1,0},{0,0,0,0,0,1}};        int l=len;        while(l){            if(l%2){                for(i=0;i<=5;i++)                    for(j=0;j<=5;j++)                        for(k=0;k<=5;k++)                            tem[i][j]+=(t[i][k]*a[k][j])%mod;                for(i=0;i<=5;i++)                    for(j=0;j<=5;j++){                        t[i][j]=tem[i][j]%mod;                        tem[i][j]=0;                    }            }            for(i=0;i<=5;i++)                for(j=0;j<=5;j++)                    for(k=0;k<=5;k++)                        tem[i][j]+=(a[i][k]*a[k][j])%mod;            for(i=0;i<=5;i++)                for(j=0;j<=5;j++){                    a[i][j]=tem[i][j]%mod;                    tem[i][j]=0;                }            l/=2;        }        ll ans=1;        ll m=2;        while(len){            if(len%2)                ans=(ans*m)%mod;            m=(m*m)%mod;            len/=2;        }        ans%=mod;        printf("%I64d\n",(ans+mod*2-t[5][5]-t[4][5])%mod);    }    return 0;}