zoukankan      html  css  js  c++  java
  • hdu 5269 ZYB loves Xor I

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 538    Accepted Submission(s): 259


    Problem Description
    Memphis loves xor very musch.Now he gets an array A.The length of A is n.Now he wants to know the sum of all (lowbit(Ai xor Aj)) (i,j[1,n])
    We define that lowbit(x)=2k,k is the smallest integer satisfied ((x and 2k)>0)
    Specially,lowbit(0)=0
    Because the ans may be too big.You just need to output ans mod 998244353
     
    Input
    Multiple test cases, the first line contains an integer T(no more than 10), indicating the number of cases. Each test case contains two lines
    The first line has an integer n
    The second line has n integers A1,A2....An
    n[1,5104]Ai[0,229]
     
    Output
    For each case, the output should occupies exactly one line. The output format is Case #x: ans, here x is the data number begins at 1.
     
    Sample Input
    2
    5
    4 0 2 7 0
    5
    2 6 5 4 0
     
    Sample Output
    Case #1: 36
    Case #2: 40
     
    /*time 62ms
    by atrp
    */
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <map>
    using namespace std;
    typedef long long ll;
    const int N = 50005;
    int a[N];
    int n, forc;
    ll ans;
    int cmp(int a, int b)
    {
        return (a & (1 << forc)) < (b & (1 << forc));
    }
    int calc(int low, int high)//寻找排序后a[low。。high - 1]中二进制第forc位不同的分界点,区间为[low,high);
    {
        int i;
        for(i = low; i < high; ++i)
            if((a[i] & (1 << forc)) ^ (a[i + 1] & (1 << forc))) break;
            if(i == high) return i;
            else return i + 1;//注意这里的边界处理
    }
    void solve(int low, int high)
    {
        sort(a + low, a + high, cmp);
        int m = calc(low, high);
        // printf("[%d] - [%d]
    ", m, high);
        int mi = *min_element(a + low, a + high);
        int mx = *max_element(a + low, a + high);
        if(mi == mx) return;//当[low,high)中的元素相等时,无需继续递归
        forc++;
        solve(low, m);
        solve(m, high);
        forc--;
        ans += ((m - low) % 998244353) * ((high - m) % 998244353) * (1 << forc) ;
    }
    int main()
    {
        int t, ca = 1;
        scanf("%d", &t);
        while(t --)
        {
            scanf("%d", &n);
            for(int i = 0; i < n; ++i) scanf("%d", &a[i]);
            forc = 0;
            ans = 0;
            solve(0, n);
            printf("Case #%d: %lld
    ", ca++,(ans << 1) % 998244353);
        }
    }
    

      

    Source
  • 相关阅读:
    【原创】QTP中手动添加对象
    【转载】【缺陷预防技术】流程技术预防
    【资料】HP Loadrunner 11下载地址
    使用命令行操作VSS
    sql server 按时间段查询记录的注意事项
    Asp.net应用程序文件名重名引起的bug
    使用SQL语句查询表中重复记录并删除
    backgroundpositionx的兼容性问题
    关于Asp.net Development Server
    如何查看正在使用某个端口的应该程序
  • 原文地址:https://www.cnblogs.com/orchidzjl/p/4609707.html
Copyright © 2011-2022 走看看