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Bitwise Tuples CodeChef Solution
Problem -Bitwise Tuples CodeChef Solution
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Bitwise Tuples CodeChef Solution in C++17
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define mod 1000000007
int fast_mutiplier(int a, int b)
{
int result = 1;
int a_p = a%mod;
while(b!=0)
{
if(b%2==1)
{
result = ((result%mod)*a_p)%mod;
}
a_p = (a_p*a_p)%mod;
b = b/2;
}
return result;
}
void solve()
{
int n, m;
cin>>n>>m;
int a = fast_mutiplier(2,n);
int b = fast_mutiplier(a-1,m);
cout<<b%mod<<endl;
}
int32_t main()
{
ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int t=1;
cin>>t;
for(int i=1; i<=t; ++i)
{
solve();
}
return 0;
}Bitwise Tuples CodeChef Solution in C++14
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define Time cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
#define pb push_back
#define mp make_pair
#define line cout << endl;
#define ff first
#define ss second
#define vi vector<int>
#define no cout << "NO" << endl;
#define yes cout << "YES" << endl;
#define printv(v) \
for (int i = 0; i < (v.size()); i++) \
{ \
cout << v[i] << " "; \
} \
line;
#define onesbits(x) __builtin_popcountll(x)
#define zerobits(x) __builtin_ctzll(x)
#define sp(x, y) fixed << setprecision(y) << x
#define w(x) \
int x; \
cin >> x; \
while (x--)
#define tk(x) \
int x; \
cin >> x;
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#ifndef ONLINE_JUDGE
#define debug(x) \
cerr << #x << " "; \
_print(x); \
cerr << endl;
#else
#define debug(x)
#endif
template <class T>
void _print(T t)
{
cerr << t;
}
template <class T, class V>
void _print(pair<T, V> p)
{
cerr << "{";
_print(p.ff);
cerr << ",";
_print(p.ss);
cerr << "}";
}
template <class T>
void _print(vector<T> v)
{
cerr << "[ ";
for (T i : v)
{
_print(i);
cerr << " ";
}
cerr << "]";
}
template <class T>
void _print(vector<vector<T>> v)
{
cerr << "[\n";
for (int l = 0; l < v.size(); l++)
{
{
for (int k = 0; k < v[l].size(); k++)
cerr << v[l][k] << " ";
}
cerr << "\n";
}
cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v)
{
cerr << "[ ";
for (auto i : v)
{
_print(i);
cerr << " ";
}
cerr << "]";
}
template <class T>
void _print(set<T> v)
{
cerr << "[ ";
for (T i : v)
{
_print(i);
cerr << " ";
}
cerr << "]";
}
const long long inf = 1e18;
const int MOD = 1e9 + 7;
const int MAX = 1e6;
int numbit(int x)
{
int ans = 0;
while (x > 0)
{
x = x >> 1;
ans++;
}
return ans;
}
int setbit(int x)
{
int ans = 0;
while (x > 0)
{
if (x & 1)
{
ans++;
}
x = x >> 1;
}
return ans;
}
bool isValid(string s)
{
int len = s.size();
for (int i = 0; i < len / 2; i++)
{
if (s[i] != s[len - 1 - i])
return false;
}
return true;
}
void rotateMatrix(vector<vector<int>> &v, int n)
{
for (int i = 0; i < n / 2; i++)
{
for (int j = i; j < n - i - 1; j++)
{
int ptr = v[i][j];
v[i][j] = v[n - 1 - j][i];
v[n - 1 - j][i] = v[n - 1 - i][n - 1 - j];
v[n - 1 - i][n - 1 - j] = v[j][n - 1 - i];
v[j][n - 1 - i] = ptr;
}
}
}
vector<bool> is_prime(10001, 1);
vector<int> primes;
void seive()
{
is_prime[0] = 0;
is_prime[1] = 0;
for (int i = 2; i < 10001; i++)
{
if (is_prime[i])
{
primes.push_back(i);
for (int j = i + i; j < 10001; j += i)
{
is_prime[j] = 0;
}
}
}
}
ll expo (ll x,ll n){
ll result=1;
while (n) {
if (n & 1)
result = result * x % MOD;
n = n / 2;
x = x * x % MOD;
}
return result;
}
int32_t main() {
ll t; cin >> t;
// seive();
while (t--) {
ll n,m;
cin>>n>>m;
ll ans=expo(2,n)-1;
cout<<expo(ans,m)<<endl;
}
return 0;
}
// 1b 4dBitwise Tuples CodeChef Solution in PYTH 3
# cook your dish here
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
for _ in range(int(input())):
[N,M]=[int(i) for i in input().split(" ")]
ans=power((power(2,N,1000000007)-1),M,1000000007)
print(ans)Bitwise Tuples CodeChef Solution in C
#include <stdio.h>
int main()
{
int x;
scanf("%d",&x);
while(x--)
{
long long int N,M;
long long int _=1;
scanf("%lld %lld",&N,&M);
long long int ans = power(2, N);
ans=ans-1;
long long int res = power(ans, M);
printf("%lld\n", res);
}
}
int power(long long int a,long long int b)
{
long long int res=1,mod=1000000007;
while(b>0){
if(b & 1)
res = (res*a%mod);
a = (a*a)%mod;
b = b>>1;
}
return res;
}Bitwise Tuples CodeChef Solution in JAVA
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
static long power(long x, long y, long p){
long res = 1L; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0L)
return 0L; // In case x is divisible by p;
while (y > 0L)
{
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
public static void main (String[] args) throws java.lang.Exception
{
int T;
Scanner in = new Scanner(System.in);
T=in.nextInt();
// in.nextLine();
while(T-->0)
{
int mod=(int)(Math.pow(10,9)+7);
int n=in.nextInt();
int m=in.nextInt();
long pow=power(2,n,mod);
pow=(pow-1);
// System.out.println(pow);
long ans=power(pow,m,mod);
System.out.println(ans);
}
}
}Bitwise Tuples CodeChef Solution in PYPY 3
T = int(input())
for _ in range(T):
N, M = map(int, input().split())
mod = 1000000007
print(pow(pow(2,N, mod)-1, M, mod))
Bitwise Tuples CodeChef Solution in PYTH
MD = 1000000007
t = int(raw_input())
for i in range(t):
st = raw_input().split()
N = int(st[0])
M = int(st[1])
v = (pow(2,N,MD)-1)%MD
r = pow(v,M,MD)
print r
# endfor i
Bitwise Tuples CodeChef Solution in C#
using System;
using System.Linq;
public class Test
{
public static void Main()
{
int t = int.Parse(Console.ReadLine());
while(t>0)
{
t--;
long []a = Console.ReadLine().Split(' ').Select(long.Parse).ToArray();
long result = power(2,a[0]) - 1;
long finalResult = power(result,a[1]);
Console.WriteLine(finalResult);
}
}
static long power(long m , long y)
{
long result = 1;
long mod = 1000000007;
while(y > 0)
{
if((y & 1) != 0)
{
result = ((result%mod)*(m%mod))%mod;
}
y >>= 1;
m = ((m%mod)*(m%mod))%mod;
}
return result;
}
}Bitwise Tuples CodeChef Solution in GO
package main
import (
"bufio"
"bytes"
"fmt"
"os"
)
func main() {
reader := bufio.NewReader(os.Stdin)
var buf bytes.Buffer
tc := readNum(reader)
for tc > 0 {
tc--
n, m := readTwoNums(reader)
res := solve(n, m)
buf.WriteString(fmt.Sprintf("%d\n", res))
}
fmt.Print(buf.String())
}
func readInt(bytes []byte, from int, val *int) int {
i := from
sign := 1
if bytes[i] == '-' {
sign = -1
i++
}
tmp := 0
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + int(bytes[i]-'0')
i++
}
*val = tmp * sign
return i
}
func readNum(reader *bufio.Reader) (a int) {
bs, _ := reader.ReadBytes('\n')
readInt(bs, 0, &a)
return
}
func readTwoNums(reader *bufio.Reader) (a int, b int) {
res := readNNums(reader, 2)
a, b = res[0], res[1]
return
}
func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
res := readNNums(reader, 3)
a, b, c = res[0], res[1], res[2]
return
}
func readNNums(reader *bufio.Reader, n int) []int {
res := make([]int, n)
x := 0
bs, _ := reader.ReadBytes('\n')
for i := 0; i < n; i++ {
for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
x++
}
x = readInt(bs, x, &res[i])
}
return res
}
func readUint64(bytes []byte, from int, val *uint64) int {
i := from
var tmp uint64
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + uint64(bytes[i]-'0')
i++
}
*val = tmp
return i
}
const MOD = 1000000007
func pow(A, B int64) int64 {
R := int64(1)
for B > 0 {
if B&1 == 1 {
R = (R * A) % MOD
}
A = (A * A) % MOD
B >>= 1
}
return R
}
func solve(n, m int) int {
// for each bit, we can choose 0 or 1, there are total 2 ^^ n ways,
// but all 1, will cause result not 0, so, we need to subtract 1
// and there are total m bits
x := pow(2, int64(n))
x += MOD - 1
x %= MOD
return int(pow(x, int64(m)))
}Bitwise Tuples CodeChef Solution Review:
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