Xor

 

Xor

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 109, Accepted users: 92

Problem 11130 : No special judgement

Problem description

We all know that huicpc302 is good at maths, now he does a job below.

At first he got a n numbers.

Then he got a function below:

long long Cal(int a[], int n)
{
	int i, j;
	long long ret = 0;
	for(i = 0; i < n; i ++)
	{
		for(j = 0; j < n; j ++)
		{
			ret += a[i]^a[j];
		}
	}
	return ret;
}

Now he wants to know what he will get, when using the function to the calculate the n numbers.
Note that when n is too large, this function works slowly, but huicpc302 wants to get the answer as soon as possible.

Input

There are several test cases in the input, the first line of each test case is an integer n, the second line contains n(1 <= n <= 100000) integers, a₀, a₁, a₂... a_n-1. (1 <= a_i < 2¹⁹). The input ends by EOF.

Output

For each test case, output one line, the result.

Sample Input

5
41 18467 6334 26500 19169
5
15724 11478 29358 26962 24464

Sample Output

338484
352232

Judge Tips

a[i]^a[j] means a[i] Xor a[j]

Problem Source

YiYi modified from xmu oj

 

#include<stdio.h>
int a[100001][19];
int main()
{
    int n,i,j,b,dd1[20];
    __int64 sum;
    while(~scanf("%d",&n))
    {
        int t=20;
        memset(dd1,0,sizeof(dd1));
        sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&b);
            j=19;
            while(b)
            {
                a[i][j]=b%2;
                if(b%2!=0)
                    dd1[j]++;
                b=b/2;
                j--;
            }
            if(j+1<t)
                t=j+1;
        }
        __int64 x=1;
        for(i=1;i<=19-t;i++)
            x=x*2;
        for(i=t;i<=19;i++)
        {
            sum=sum+x*dd1[i]*(n-dd1[i]);
            x=x/2;
        }
        printf("%I64d\n",2*sum);
    }
    return 0;
}