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Minimise the Size CodeChef Solution
Problem -Minimise the Size CodeChef Solution
“This website is dedicated for CodeChef solution where we will publish right solution of all your favourite CodeChef problems along with detailed explanatory of different competitive programming concepts and languages.
Minimise the Size CodeChef Solution in C++17
“ #include <bits/stdc++.h>
using namespace std;
#define int long long int
signed main()
{
int t,c,p;
cin >> t;
while(t--)
{
vector <int> ans;
cin >> c;
if (c == 0){
cout << 2 << endl;
cout << 1 << " " << 1 << endl;
continue;
}
while(1)
{
p = log2(c) + 1;
p = pow(2,p);
p--;
ans.push_back(p);
if (p == c)
{
break;
}
c = p - c;
}
cout << ans.size() << endl;
for(auto a : ans) cout << a << " ";
cout << endl;
}
return 0;
}
// check the constratins for edge cases. Minimise the Size CodeChef Solution in C++14
“#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
long long int C;
cin>>C;
vector<long long int> result;
long long int curr = 0;
for(int i=60; i>=0;i--){
if( (curr & (1LL<<i)) != (C & (1LL<<i)) )
{
long long int b = (1LL<<(i+1)) -1LL;
result.push_back(b);
curr =curr^b;
}
}
if(result.size()==0){
result.push_back(1);
result.push_back(1);
}
std::cout << result.size() << std::endl;
for(auto x : result)
cout<< x <<" ";
cout<<endl;
}
return 0;
}
Minimise the Size CodeChef Solution in PYTH 3
“t = int(input())
def solve():
C = int(input())
if C == 0:
print(2)
print(1, 1)
return
xor = 0
ans = []
for i in range(60,-1,-1):
if C & (1<<i):
if xor & (1<<i):
continue
else:
ans.append((1<<(i+1)) - 1)
xor ^= ((1<<(i+1))-1)
else:
if not xor & (1<<i):
continue
else:
ans.append((1<<(i+1))-1)
xor ^= ((1<<(i+1))-1)
print(len(ans))
if ans:
print(*ans)
else:
print(0)
while t:
solve()
t -= 1Minimise the Size CodeChef Solution in C
“#include <stdio.h>
#include <math.h>
#include <string.h>
void solve();
void getBinary(char[], long int);
void increaseByOne(int[], int, int);
void solve()
{
long int c; //c=17
scanf("%ld", &c);
if (c == 0)
printf("2\n1 1");
else
{
int len = log2(c) + 1; //len = 5
char binary[len]; //binary = 10001
getBinary(binary, c);
int ones[len], ans[len], k = 0;
for (int i = 0; i < len; i++)
ones[i] = 0;
for (int i = 0; i < len; i++)
{
if (binary[i] == '1')
{
if (ones[i] % 2 != 1)
{
increaseByOne(ones, i, len);
ans[k] = len - i;
k++;
}
}
else
{
if (ones[i] % 2 == 1)
{
increaseByOne(ones, i, len);
ans[k] = len - i;
k++;
}
}
}
printf("%d\n", k);
for (int i = k - 1; i >= 0; i--)
{
long int final = pow(2, ans[i]);
final--;
printf("%ld ", final);
}
}
printf("\n");
}
void getBinary(char arr[], long int num)
{
if (num == 0)
{
arr[0] = '0';
return;
}
int last = log2(num);
while (num)
{
int rem = num % 2;
arr[last--] = rem ? '1' : '0';
num /= 2;
}
}
void increaseByOne(int arr[], int start, int end)
{
for (int i = start; i < end; i++)
{
arr[i]++;
}
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
solve();
}
}Minimise the Size CodeChef Solution in JAVA
“
import java.io.*;
import java.util.*;
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
/* find distance of one postion wrt to another in circular arrray..
for(int i = 0; i< 2*n; i++)
time[(i+1)%n] = Math.min(time[(i+1)%n], time[i%n]+1); //checking with bacward position
for(int i = 2*n-1; i>= 0; i--)
time[(i+n-1)%n] = Math.min(time[(i+n-1)%n], time[i%n]+1);
*/
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
class Codechef {
static class pair implements Comparable<pair> {
int a;
int b;
pair(int x,int y){
a= x;
b= y;
}
public int compareTo(pair p) {
return b-p.b;
}
}
static boolean factors(int num){
for(int i=2;i*i<=num;i++) {
if(num%2==0) {
return false;
}
}
return true;
}
static long coPrimes(long n) {
long res = n;
for(int i=2;i*i<=n;i++) {
if(n%i==0) {
res = res/i;
res = res*(i-1);
while(n%i==0) {
n = n/i;
}
}
}
if(n!=1) {
res = res/n;
res = res*(n-1);
}
return res;
}
static long mod = 1000000007;
static void solve(FastReader sc, PrintWriter out) {
long c = sc.nextLong();
ArrayList<Long> al = new ArrayList<>();
long curr_xor= 0;
for(int i=60;i>=0;i--) {
long cBit = ((1l<<i) & c);
long curr_xorBit = ((1l<<i)& curr_xor);
if(cBit!=curr_xorBit) {
al.add( (1l<<(i+1))-1);
curr_xor = curr_xor^( (1l<<(i+1))-1);
}
}
if(al.size()==0) {
al.add(1l);
al.add(1l);
}
out.println(al.size());
for(int i=0;i<al.size();i++) {
out.print( al.get(i)+" ");
}
out.println();
}
static long rightMostSetBit(long n){
//will give only rmsb and all other will be zero.
return n&(-n);
}
static int largestPowerOfTwoInRange(long n) {
int x = 0;
while((1<<x)<=n) {
x++;
}
return x-1;
//log2 can also give right most set bit.
}
public static void main(String[] args) throws IOException{
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-->0){
solve(sc,out);
}
out.close();
}
}
//Math.ceil(Math.log(x)/Math.log(2)) gives X which satisfy 2^X.Minimise the Size CodeChef Solution in PYPY 3
“t = int(input())
def solve():
C = int(input())
if C == 0:
print(2)
print(1, 1)
return
xor = 0
ans = []
for i in range(60,-1,-1):
if C & (1<<i):
if xor & (1<<i):
continue
else:
ans.append((1<<(i+1)) - 1)
xor ^= ((1<<(i+1))-1)
else:
if not xor & (1<<i):
continue
else:
ans.append((1<<(i+1))-1)
xor ^= ((1<<(i+1))-1)
print(len(ans))
if ans:
print(*ans)
else:
print(0)
while t:
solve()
t -= 1Minimise the Size CodeChef Solution in PYTH
“L = [1]
v = 1
for k in range(59):
v = v*2 +1
L.append(v)
# endfor k
L.reverse()
t = int(raw_input())
for i in range(t):
C = int(raw_input())
if C == 0:
print '2'
print '1 1'
else:
B = []
R = []
for k in range(60):
B.append(C%2)
C = C/2
# endfor k
B.reverse()
n = 0
for k in range(60):
if (B[k]+n)%2 == 1:
R.append(L[k])
n += 1
# endif
# endfor k
print len(R)
st = ''
for x in R:
st += str(x) + ' '
# endfor x
print st
# endif
# endfor iMinimise the Size CodeChef Solution in GO
“package main
import (
"bufio"
"bytes"
"fmt"
"os"
)
func main() {
reader := bufio.NewReader(os.Stdin)
tc := readNum(reader)
var buf bytes.Buffer
for tc > 0 {
tc--
var C uint64
s, _ := reader.ReadBytes('\n')
readUint64(s, 0, &C)
res := solve(C)
buf.WriteString(fmt.Sprintf("%d\n", len(res)))
for i := 0; i < len(res); i++ {
buf.WriteString(fmt.Sprintf("%d ", res[i]))
}
buf.WriteByte('\n')
}
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
}
func solve(C uint64) []uint64 {
if C == 0 {
return []uint64{1, 1}
}
var res []uint64
for i := uint64(60); ; i-- {
if (C>>i)&1 == 1 {
if len(res)%2 == 0 {
res = append(res, (1<<(i+1) - 1))
}
} else {
if len(res)%2 == 1 {
res = append(res, (1<<(i+1) - 1))
}
}
if i == 0 {
break
}
}
return res
}Minimise the Size CodeChef Solution Review:
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