1615A - Closing The Gap - CodeForces Solution


greedy math *800

Please click on ads to support us..

Python Code:

import math
t=int(input())
for zz in range(t):
    n=int(input())
    l=list(map(int,input().split()))
    s=sum(l)
    if s%n==0:
        print(0)
    else:
        print(1)
        
        
    

C++ Code:

#include<bits/stdc++.h>
#include <iostream>
#define ll long long 
#include <algorithm>
#define int long long int
#define take(x) for(auto& a:x)cin>>a
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define arrmin(x) *min_element(all(x))
#define arrmax(x) *max_element(all(x))
#define arrsum(x) accumulate(all(x), 0LL)
#define arrpre(x) partial_sum(all(x), x.begin())
#define arrsuf(x) partial_sum(allr(x), x.rbegin())
#define give(x) for(auto a:x){cout<<a<<' ';}cout<<endl
#define PI 3.1415926535897932384626
#define F              first
#define S              second
#define pb             push_back
#define si             set <int>
#define vi             vector <int>
#define pii            pair <int, int>
#define vpi            vector <pii>
#define vpp            vector <pair<int, pii>>
#define mii            map <int, int>
#define mpi            map <pii, int>
#define spi            set <pii>
#define double         long double
#define que_max        priority_queue <int>
#define que_min        priority_queue <int, vi, greater<int>>
#define mini  *min_element(a,a + n);
#define maxi  *max_element(a,a + n);
#define maxis *max_element(s.begin(),s.end());
#define minis *min_element(s.begin(),s.end());
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define ppc(x) __builtin_popcount(x);
#define ppcl(x) __builtin_popcountll(x);
#define clz(x) __builtin_clz(x); // // Leading zeroes count
#define ctz(x) __builtin_ctz(x); // Trailing zeroes count
#define nline endl;
#define print(x) cout << x << endl
#define mod 1000000007
using namespace std;
 
 
 
 
int gcd( int n1, int n2){  
    if(n2==0)
        return n1;
    return gcd(n2,n1%n2);
}  
int lcm (int n1, int n2)  {    
 
return  (n1 / gcd(n1, n2)) * n2;
 
}



int power(int a, int n)
{
 
    // Stores final answer
    int ans = 1;
 
    while (n > 0) {
 
        int last_bit = (n & 1);
 
        // Check if current LSB
        // is set
        if (last_bit) {
            ans = ans * a;
        }
 
        a = a * a;
 
        // Right shift
        n = n >> 1;
    }
 
    return ans;
}
   
 
int no_of_digits(int n)
{
    return (floor(log10(n)) + 1);
}
 
bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to square root of n
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
 
int largestPower(int n, int p)
{
    // Initialize result
    int x = 0;
 
    // Calculate x = n/p + n/(p^2) + n/(p^3) + ....
    while (n) {
        n /= p;
        x += n;
    }
    return x;
}
 
 
 
int mex(int n)
{
    int k=0;
    if(k!=n)return k;
    return k+1;
}
 
bool sortbysecdesc(const pair<int,int> &a,
        const pair<int,int> &b)
{
 
    if(a.first == b.first){
        return a.second <b.second;
    }
    else
    return a.first>b.first;
}
 
int BinarySearchUminus(int val, vector<int> &A)
{
    int low = 0 ,high = A.size()-1;
 
    while (high>low)
    {
        int mid = (low + high+1)/2;
        if(A[mid] == val)
        low = mid; // high = mid-1 for lower_bound-1
        else
        if(A[mid]<val)
        low = mid;
        else
        high = mid -1;
 
    }
    if(A[low] > val)
    return -1;
    return low;
    
}
 
int number_of_factors_sqrt_time(int x)
{
    vector <int> result;
    int i = 1;
    // This will loop from 1 to int(sqrt(x))
    while(i*i <= x) {
        // Check if i divides x without leaving a remainder
        if(x % i == 0) {
            result.push_back(i);
            // Handle the case explained in the 4th
            if(x/i != i) {
                result.push_back(x/i);
            }
        }
        i++;
    }
    return result.size();
}
 
int palindrome(vector<int> arr)
{
    // Initialise flag to zero.
    int flag = 0;
 
    int n = arr.size();
 
    // Loop till array size n/2.
    for (int i = 0; i <= n / 2 && n != 0; i++) {
 
        // Check if first and last element are different
        // Then set flag to 1.
        if (arr[i] != arr[n - i - 1]) {
            flag = 1;
            break;
        }
    }
 
    if (flag == 1)
        return 0;
    else
        return 1;
}
 
int highestPoweroftwo(int n)
{
    int p = (int)log2(n);
    return (int)pow(2, p);
    
    // int res = 0;
    // for (int i = n; i >=1; i--)
    // {
    //     /* code */
    //     if((i&(i-1)) == 0)
    //     {
    //         res = i;
    //         break;
    //     }
    // }
    // return res;
}
 
 
int ceil(int a , int b)
{
    int ans = a/b;
    if(a%b!=0)
    ans++;
    return ans;
}
 
bool compare(pair<int,int>a,pair<int,int>b){
//sort  pairs in decreasing order of the second elements
//of the pairs.If both the second elements are equal sort
//according to increasing order of the first elements of the pairs.
    if(a.second==b.second)
    return a.first<b.first;
    return a.second>b.second;
}
 
 
 
 
 
 
 
 bool ispossible(int a[],int n,int mid,int k)
 {
    int cow = 1;
    int last_pos = a[0];

    for(int i=0;i<n;i++)
    {
        
        if(a[i] - last_pos >= mid)
        {
            cow ++;
            
            if(cow == k)
                return true;
            
            last_pos = a[i];
        }
        
    }
    
  
    return false;
 }
 
 
 int min(int x,int y)
 {
    if(x>=y) return y;
    
    return x;
    
 }




int fact(int n)
{
    if(n==1)
    {
        return 1;
    }
    
    
    return n*fact(n-1);
}



// 1st element by increasing second element by increasing
bool sortbyCond(const pair<int, int>& a,
                const pair<int, int>& b)
{
    if (a.first != b.first)
        return (a.first < b.first);
    else
       return (a.second > b.second);
}


string multiply(string num1, string num2)
{
    int len1 = num1.size();
    int len2 = num2.size();
    if (len1 == 0 || len2 == 0)
    return "0";
    vector<int> result(len1 + len2, 0);
    int i_n1 = 0;
    int i_n2 = 0;
    for (int i=len1-1; i>=0; i--)
    {
        int carry = 0;
        int n1 = num1[i] - '0';
        i_n2 = 0;        
        for (int j=len2-1; j>=0; j--)
        {
            int n2 = num2[j] - '0';
            int sum = n1*n2 + result[i_n1 + i_n2] + carry;
            carry = sum/10;
            result[i_n1 + i_n2] = sum % 10;
            i_n2++;
        }
        if (carry > 0)
            result[i_n1 + i_n2] += carry;
         i_n1++;
    }
    int i = result.size() - 1;
    while (i>=0 && result[i] == 0)
    i--;
    if (i == -1)
    return "0";
    string s = "";  
    while (i >= 0)
        s += std::to_string(result[i--]);
    return s;
}


string findDiff(string str1, string str2)
{
    string str = "";
    int n1 = str1.length(), n2 = str2.length();
    int diff = n1 - n2;
    int carry = 0;
    for (int i = n2 - 1; i >= 0; i--) {
        int sub = ((str1[i + diff] - '0') - (str2[i] - '0')
                   - carry);
        if (sub < 0) {
            sub = sub + 10;
            carry = 1;
        }
        else
            carry = 0;
        str.push_back(sub + '0');
    }
    for (int i = n1 - n2 - 1; i >= 0; i--) {
        if (str1[i] == '0' && carry) {
            str.push_back('9');
            continue;
        }
        int sub = ((str1[i] - '0') - carry);
        if (i > 0 || sub > 0) // remove preceding 0's
            str.push_back(sub + '0');
        carry = 0;
    }
    reverse(str.begin(), str.end());
    return str;
}

int msbPos(ll n)
{
    int msb_p = 0;
    while (n)
    {
        n = n>>1;
        msb_p++;
    }
    return msb_p;
}

// Function to find Bit-wise & of all numbers from x
// to y.
ll andOperator(ll x, ll y)
{
    ll res = 0; // Initialize result
 
    while (x && y)
    {
        // Find positi ons of MSB in x and y
        int msb_p1 = msbPos(x);
        int msb_p2 = msbPos(y);
 
        // If positions are not same, return
        if (msb_p1 != msb_p2)
            break;
 
        // Add 2^msb_p1 to result
        ll msb_val =  (1 << msb_p1);
        res = res + msb_val;
 
        // subtract 2^msb_p1 from x and y.
        x = x - msb_val;
        y = y - msb_val;
    }
 
    return res;
}


string convert(int &n)
{
    return to_string(n);
}

int convert1(string s)
{
    return stoll(s);
}

int modInverse(int a){
    return power(a,mod-2);
}



int powermod(int a,int b) {
    int res=1;
    a%=mod;
     assert(b>=0);
 for(;b;b>>=1){
    if(b&1)
    res=res*a%mod;
    a=a*a%mod;
    }
    return res;
}




void driver()
{

int n; cin >> n;

vi a(n);
take(a);

int x = arrsum(a);

if(x%n==0)
{
    cout << 0 << endl;
    return;
}

cout << 1 << endl;
}

 





 


 

   

 
 

 

 
int32_t main()
{
    
    ios_base::
    sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    
    #ifndef ONLINE_JUDGE
    freopen("input1.txt" , "r" ,stdin);
    freopen("output1.txt" , "w" ,stdout);
    #endif
    
    clock_t z = clock();
 
 
  int t;
    //int t = 1;
cin >> t;
while(t--)
    {
        


       driver();

   }
    cerr << "Run Time : " << ((double)(clock() - z) / CLOCKS_PER_SEC);
     
} 
 
 


Comments

Submit
0 Comments
More Questions

580A- Kefa and First Steps
1472B- Fair Division
996A - Hit the Lottery
MSNSADM1 Football
MATCHES Playing with Matches
HRDSEQ Hard Sequence
DRCHEF Doctor Chef
559. Maximum Depth of N-ary Tree
821. Shortest Distance to a Character
1441. Build an Array With Stack Operations
1356. Sort Integers by The Number of 1 Bits
922. Sort Array By Parity II
344. Reverse String
1047. Remove All Adjacent Duplicates In String
977. Squares of a Sorted Array
852. Peak Index in a Mountain Array
461. Hamming Distance
1748. Sum of Unique Elements
897. Increasing Order Search Tree
905. Sort Array By Parity
1351. Count Negative Numbers in a Sorted Matrix
617. Merge Two Binary Trees
1450. Number of Students Doing Homework at a Given Time
700. Search in a Binary Search Tree
590. N-ary Tree Postorder Traversal
589. N-ary Tree Preorder Traversal
1299. Replace Elements with Greatest Element on Right Side
1768. Merge Strings Alternately
561. Array Partition I
1374. Generate a String With Characters That Have Odd Counts