zoukankan      html  css  js  c++  java
  • poj3734 Blocks[矩阵优化dp or 组合数学]

    Blocks
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 6578   Accepted: 3171

    Description

    Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.

    Input

    The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.

    Output

    For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.

    Sample Input

    2
    1
    2

    Sample Output

    2
    6

    Source

    方法1:
    //f[i]=6*f[i-1]-8*f[i-2]{i>=3,f[1]=2,f[2]=6}
    #include<cstdio>
    #include<cstring>
    typedef long long ll;
    using namespace std;
    const ll mod=10007;
    struct matrix{
        ll s[2][2];
        matrix(){
            memset(s,0,sizeof s);
        }
    }A,F;int n,T;
    matrix operator *(const matrix &a,const matrix &b){
        matrix c;
        for(int i=0;i<2;i++){
            for(int j=0;j<2;j++){
                for(int k=0;k<2;k++){
                    c.s[i][j]+=a.s[i][k]*b.s[k][j];
                    c.s[i][j]%=mod;
                }
            }
        }
        return c;
    }
    matrix fpow(matrix a,int p){
        matrix res;
        for(int i=0;i<2;i++) res.s[i][i]=1;
        for(;p;p>>=1,a=a*a) if(p&1) res=res*a;
        return res;
    }
    int main(){
        for(scanf("%d",&T);T--;){
            scanf("%d",&n);
            if(n==1){puts("2");continue;}
            if(n==2){puts("6");continue;}
            A.s[0][0]=6;A.s[0][1]=-8;
            A.s[1][0]=1;A.s[1][1]=0;
            F.s[0][0]=6;F.s[0][1]=0;
            F.s[1][0]=2;F.s[1][1]=0;
            A=fpow(A,n-2);
            F=A*F;
            printf("%lld
    ",(F.s[0][0]+mod)%mod);
        }
        return 0;    
    }

    方法2:

    //f(n)=2^(2n-2)+2^(n-1)
    #include<cstdio>
    #include<cstring>
    typedef long long ll;
    using namespace std;
    const ll mod=10007;
    ll ans=0;int T,n;
    ll fpow(ll a,ll p){
        ll res=1;
        for(;p;p>>=1,a=a*a%mod) if(p&1) res=res*a%mod;
        return res;
    }
    int main(){
        for(scanf("%d",&T);T--;){
            scanf("%d",&n);
            ans=fpow(2,n-1<<1)+fpow(2,n-1);
            printf("%I64d
    ",(ans+mod)%mod);
        }
        return 0;    
    }
  • 相关阅读:
    paramiko模块
    linux 文件权限管理
    itext 生成pdf文件添加页眉页脚
    Python3 断言
    net core WebApi——文件分片上传与跨域请求处理
    Linux配置部署_新手向(二)——Nginx安装与配置
    Linux配置部署_新手向(一)——CentOS系统安装
    net core Webapi基础工程搭建(七)——小试AOP及常规测试_Part 2
    net core Webapi基础工程搭建(六)——数据库操作_Part 2
    net core Webapi基础工程搭建(七)——小试AOP及常规测试_Part 1
  • 原文地址:https://www.cnblogs.com/shenben/p/6725996.html
Copyright © 2011-2022 走看看