Skip to main content

SCALE - Funny scales

Problem Link
#include <iostream>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;

const int N = 1e4;
int rhs[N];
int lhs[N];


int convertToTernary(long long x, int arr[], int index)
{
  if(x == 0)
  return index;
  int remainder = x % 3;
  arr[index] = remainder;
  x = x / 3;

  return convertToTernary(x, arr, index + 1);
}

long convertToDecimal(int arr[], int len)
{
long result = 0;
long power = 1;
int i = 0;
while(i < len)
{
result += arr[i] * power;
power *= 3;
i++;
}

return result;

}

void solve(int arr[],  int len, long maxSum)
{
int i = 0;
int carry = 0;
int rlen = 0;
while(i < len)
{
arr[i] += carry;
carry = arr[i] / 3;
arr[i] = arr[i] % 3;

if(arr[i] == 0 || arr[i] == 1)
{
lhs[i] = 0;
}
else
{
lhs[i] = 1;
carry = 1;
}

rhs[i] = (arr[i] + lhs[i]) % 3;
i++;
}

rlen = len;
if(carry)
{
rhs[i] = carry;
rlen++;
}

long r = convertToDecimal(rhs, rlen);

// there is a chance of overflow in rhs, so we must check
//if it is greater than maxSum or not.
if(r > maxSum)
{
cout << "-1" << endl;
return;
}

for(int i = 0; i < len; i++)
{
if(lhs[i])
{
cout << i + 1 << " ";
}
}

cout << endl;

for(int i = 0; i < rlen; i++)
{
if(rhs[i])
{
cout << i + 1 << " ";
}
}

cout << endl;

}

int main()
{

int n;
long long x;

cin >> n >> x;

int arr[N];
memset(arr, 0, sizeof arr);
memset(lhs, 0, sizeof lhs);
memset(rhs, 0, sizeof rhs);

if(x == 0 || x == 1){
cout << "1" << endl;
return 0;
}

// calculate gp series of 1, 3, 9, 27, ...
//formula = (a(r^n - 1)) / (r - 1)
// a = 1, r = 3
long maxSum = ((pow(3, n) - 1)) / 2;
if(x > maxSum)
{
cout << "-1" << endl;
return 0;
}
int len = convertToTernary(x, arr, 0);

solve(arr, len, maxSum);
return 0;
}

Comments

Post a Comment

Popular posts from this blog

MAIN72

MAIN72 - Subset sum #include <iostream> #include <cstring> using namespace std; bool dp[100001][1001]; int arr[1001]; int main() {     int t, n;     long long int sum;     cin >> t;     while(t--)     {         cin >> n;         memset(dp, false, sizeof(dp));         sum = 0;         for(int i = 0; i < n; i++)         {             cin >> arr[i];             sum += arr[i];         }         for(int i = 0; i < n; i++)             dp[0][i] = true; // 0 sum         for(int i = 1; i < n; i++)             dp[i][0] = false; // sum is i but 0 element         for(long int i = 1; i <= sum; i++)         {             for(int j = 1; j <= n; j ++)             {                 dp[i][j] = dp[i][j - 1];                 if(i >= arr[j - 1])                     dp[i][j] = dp[i][j] || dp[i - arr[j - 1]][j - 1];             }         }         long int result = 0;         for(int i = 1; i <= sum; i++)  

War of XORs- XORIER

Problem Link #include <iostream> using namespace std; int main() { int t, n, odd, even; cin >> t; while(t--) { cin >> n; int i,arr[n],freq[1100001]={0}; long res = 0; odd = even = 0; for(int i = 0; i < n; i++) { cin >> arr[i]; freq[arr[i]]++; } for(int i = 0; i < n; i++) { if(arr[i] & 1) { odd++; } else { even++; } } for(int i = 0; i < n; i++) { if(arr[i] % 2) { res += odd; } else { res += even; } res -= freq[arr[i] ^ 2]; res -= freq[arr[i]]; } cout << res / 2 << endl; } }

GCJ101BB - Picking Up Chicks

Problem Link /* explanation     lets solve the problem only for 2 chicken.     s[i] = speed of chicken i     pos[i] = position of chicken i     if s[i] > s[i - 1] then no problem, just check whether both can reach b within time or not.     if s[i] < s[i - 1] then there is a chance that i can slow down i - 1.     lets say s[i] = 1 m/sec and s[i - 1] = 2 m/sec and time limit is T and point to reach is B.     for s[i] pos[i] can be at max B - T. if pos is greater than B-T it can not reach within Tsec.     and for s[i - 1] pos[i - 1] can be at max (B-T)*2. if pos[i - 1] > (B-T)*2 it can not reach within Tsec.     at T sec i will be at B and i - 1 will also be at B. at T - 1 i will be at B-T-1 and i-1 will be at B-T-2 and so on. as we can see i -1 will always be behind i. so there will not be any collision.     if i is pos[i] < B-T then i can reach B before T sec and it will not cause any problem .     problem may occur if pos[i - 1]< (B -T)*2 as it can me