Xoring Ninja
Problem Statement :
An XOR operation on a list is defined here as the xor () of all its elements (e.g.: ). The of set is defined here as the sum of the s of all non-empty subsets of known as . The set can be expressed as: For example: Given set The set of possible non-empty subsets is: The of these non-empty subsets is then calculated as follows: = Given a list of space-separated integers, determine and print . For example, . There are three possible subsets, . The XOR of , of and of . The XorSum is the sum of these: and . Note: The cardinality of powerset is , so the set of non-empty subsets of set of size contains subsets. Function Description Complete the xoringNinja function in the editor below. It should return an integer that represents the XorSum of the input array, modulo . xoringNinja has the following parameter(s): arr: an integer array Input Format The first line contains an integer , the number of test cases. Each test case consists of two lines: - The first line contains an integer , the size of the set . - The second line contains space-separated integers . Output Format For each test case, print its on a new line. The line should contain the output for the test case.
Solution :
Solution in C :
In C :
#include<stdio.h>
#define LL long long int
#define MOD 1000000007
LL power(LL a, LL b)
{
if (b == 0)
return 1;
else
{
LL temp=(power(a,b/2))%MOD;
if(b%2==0)
return (temp*temp)%MOD;
else
return (((temp*a)%MOD)*temp)%MOD;
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
int x=0;
int i,a;
for(i=0;i<n;i++)
{
scanf("%d",&a);
x|=a;
}
LL ans=power(2,n-1);
ans=(ans*x)%MOD;
printf("%lld\n",ans);
}
return 0;
}
Solution in C++ :
In C++ :
#include <cstdio>
#include <iostream>
#include <fstream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <ctime>
#include <cassert>
#include <utility>
using namespace std;
#define MAXN 100005
#define MOD 1000000007
int T, N;
int A[MAXN];
long long dp[MAXN][2];
int main() {
// freopen("date.in", "r", stdin);
cin >> T;
while(T--) {
cin >> N;
for(int i = 0; i < N; i++)
cin >> A[i];
long long ans = 0;
for(int bit = 0; bit < 32; bit++) {
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
dp[0][1] = 0;
for(int i = 0; i < N; i++) {
int crt = 0;
if(A[i] & (1 << bit))
crt = 1;
for(int j = 0; j < 2; j++) {
int nj = j ^ crt;
dp[i + 1][nj] += dp[i][j];
dp[i + 1][j] += dp[i][j];
dp[i + 1][nj] %= MOD;
dp[i + 1][j] %= MOD;
}
}
long long cnt = dp[N][1];
ans += cnt * (1LL << bit);
ans %= MOD;
}
cout << ans << '\n';
}
return 0;
}
Solution in Java :
In Java :
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class Solution {
static Solution main;
static long sp;
public static void main(String[] args) {
sp = 1000000007;
long [] fact = new long[110001];
long [] pow2 = new long[110001];
pow2[0] = 1;
fact[0] = 1;
for(int i = 1 ; i < 110001 ; i++) {
fact[i] = fact[i-1] * i;
fact[i] %= sp;
pow2[i] = 2*pow2[i-1];
if(pow2[i]>=sp) {
pow2[i]-=sp;
}
}
main = new Solution();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedOutputStream bos = new BufferedOutputStream(System.out);
String eol = System.getProperty("line.separator");
byte[] eolb = eol.getBytes();
try {
String str = br.readLine();
int t = Integer.parseInt(str);
for(int i = 0 ; i < t ; i++) {
str = br.readLine();
int n = Integer.parseInt(str);
int [] ar = new int[40];
Arrays.fill(ar, 0);
str = br.readLine();
int j=0;
int s=0;
int length = str.length();
while(j<length) {
while(j<length) {
if(str.charAt(j) == ' ') {
break;
}else {
j++;
}
}
int x = Integer.parseInt(str.substring(s,j)) ;
int iter = 0;
while(x>0) {
if((x%2)==1) {
ar[iter]++;
}
x/=2;
iter++;
}
j++;
s=j;
}
long ans = 0;
for(int a = 0 ;a < 40 ; a++) {
int x = ar[a];
for(int b = 1 ; b <= x; b+=2) {
// x choose b * 2^(n-x) * 2^a
long tempAns = fact[x];
tempAns *= inverse(fact[b]);
tempAns %= sp;
tempAns *= inverse(fact[x-b]);
tempAns %= sp;
tempAns *= pow2[n-x+a];
tempAns %= sp;
ans += tempAns;
if(ans>sp) {
ans -= sp;
}
}
}
bos.write(new Long(ans).toString().getBytes());
bos.write(eolb);
}
bos.flush();
} catch(IOException ioe) {
ioe.printStackTrace();
}
}
public static long inverse(long a) {
return getPow(a,sp-2);
}
public static long getPow(long a , long b) {
long ans = 1;
long pow = a;
while(b>0) {
if((b%2)==1) {
ans *= pow;
ans %= sp;
}
pow *= pow;
pow %= sp;
b/=2;
}
return ans;
}
}
Solution in Python :
In Python3 :
import sys
def poweroftwo(n):
val = 1
yield val
for i in range(n):
val *= 2
yield val
r = sys.stdin.readline
#table = list(poweroftwo(100000))
text = r()
t = int(text)
for i in range(t):
text = r()
n = int(text)
text = r()
tmp = 0
for num in map(int,text.split()):
tmp |= num
#tmp *= table[n]
for i in range(n-1):
tmp *= 2
if i % 20 == 0:
tmp %= 1000000007
print(tmp%1000000007)
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