zoukankan      html  css  js  c++  java
  • codeforces 691E Xor-sequences(矩阵快速幂)

    引自:http://www.cnblogs.com/shuguangzw/p/5674089.html

    /* ***********************************************
    Author        :devil
    ************************************************ */
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <cmath>
    #include <stdlib.h>
    using namespace std;
    typedef long long LL;
    const int inf=0x3f3f3f3f;
    const int mod=1e9+7;
    const int N=1e2+10;
    LL a[N],k,ans;
    int n;
    struct Matrix
    {
        LL a[N][N];
        Matrix()
        {
            memset(a,0,sizeof(a));
        }
        void init()
        {
            for(int i=1;i<=n;i++)
                a[i][i]=1;
        }
        Matrix operator *(const Matrix &rhs)const
        {
            Matrix ret;
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)
                    for(int k=1;k<=n;k++)
                        ret.a[i][j]=(ret.a[i][j]+a[i][k]*rhs.a[k][j]%mod)%mod;
            return ret;
        }
        Matrix operator ^(LL mi)const
        {
            Matrix tmp=(*this),ret;
            ret.init();
            while(mi)
            {
                if(mi&1)ret=ret*tmp;
                tmp=tmp*tmp;
                mi>>=1;
            }
            return ret;
        }
    }cur;
    int main()
    {
        //freopen("in.txt","r",stdin);
        scanf("%d%I64d",&n,&k);
        for(int i=1; i<=n; i++)
            scanf("%I64d",&a[i]);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(__builtin_popcountll(a[i]^a[j])%3==0)
                    cur.a[i][j]++;
        cur=cur^(k-1);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                ans=(ans+cur.a[i][j])%mod;
        printf("%I64d
    ",ans);
        return 0;
    }
  • 相关阅读:
    计算中文混合字符串长度(一)
    PHP截取含中文的混合字符串长度的函数
    获取星座的JS函数
    获取生日对应星座的PHP函数
    简单的 jQuery 浮动层随窗口滚动滑动插件实例
    MD5算法实现
    70. Climbing Stairs QuestionEditorial Solution
    167. Two Sum II
    167. Two Sum II
    303. Range Sum Query
  • 原文地址:https://www.cnblogs.com/d-e-v-i-l/p/5697170.html
Copyright © 2011-2022 走看看