Hug The Princess (thinking, bitwise arithmetic)

Hug the PrincessTime limit:3000/1000ms (java/others) Memory limit:65535/65535kb (java/others)Submit Status

There is a sequence withNNElements. Assuming they is a1,a2,?,an a1,a2,?, an.

Please calculate the following expession.

∑1≤I<J≤N(AI∧AJ)+ (a i| aj ) + (a I&aj) ∑1≤i<j≤n (Ai∧aj) + (Ai|aj) + (Ai&aj)

In the expression above, ^  |  &  is bit operation. If you don ' t know bit operation, you can visit

Http://en.wikipedia.org/wiki/Bitwise_operation

To get some useful information.

Input

The first line contains a single integer nN and which is the size of the sequence.

The second line containsNNIntegers, the ItH ithIntegerai ai is the   i t h "> ith ith element of the sequence.

1≤n≤100000,0≤ai≤100000000 1≤n≤100000,0≤ai≤100000000

Output

Print the answer in one line.

Sample Input and output

Sample Input	Sample Output
21 2	6

Hint

Because the answer is so large, please use a long long instead of int. Correspondingly, please use %lld instead of< c3> to %d scanf and printf.

Large input. You could get time Limit exceeded if you use "cin" to get the input. So "scanf" is suggested.

Likewise, you is supposed to use "printf" instead of "cout".

Exercises

Let this expression, we can find each bit of the prefix and then based on each bit to figure out the front is 1, is 0 of the current number of bits, and then add CNT * (1 << j) just fine.

Code:

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intMAXN = 1e5 + -;intnum[maxn][ *];inta[ *];intP[maxn];typedefLong LongLL;intMain () {intN;  while(~SCANF ("%d", &N)) {        intx; memset (num,0,sizeof(num));  for(inti =1; I <= N; i++) {scanf ("%d", &x);  for(intj =0; J < -; J + +) {Num[i][j]= Num[i-1][J] + x%2; X= x/2; }} LL ans=0;  for(inti =1; I <= N; i++){            inttemp =0, pos =-1;  for(intj =0; J < -; J + +) {A[j]= Num[i][j]-num[i-1][j]; //printf ("%d", a[j]);}//puts ("");             for(intj =0; J < -; J + +){                intCNT =0; if(A[j]) {CNT+ = (I-1-Num[i-1][j]); CNT+ = i-1; CNT+ = Num[i-1][j]; }                Else{cnt+ = Num[i-1][j]; CNT+ = Num[i-1][j]; } ans+ = (LL) CNT * (1<<j); }} printf ("%lld\n", ans); }    return 0;}

Hug The Princess (thinking, bitwise arithmetic)