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Minimise the Size CodeChef Solution
Problem -Minimise the Size CodeChef Solution
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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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