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


[align=left]Problem Description[/align]
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

[align=left]Input[/align]
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∈[1,5∗104]，Ai∈[0,229]

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2

5
4 0 2 7 0
5
2 6 5 4 0

[align=left]Sample Output[/align]

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
;
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);
}
}

　　

[align=left]Source[/align]
BestCoder Round #44