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
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∈[1,5∗104]，Ai∈[0,229]
The first line has an integer n
The second line has n integers A1,A2....An
n∈[1,5∗104]，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