hdu 5269 ZYB loves Xor I

/*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]\n", 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\n", ca++,(ans << 1) % 998244353);
    }
}