# Daily Coding Challenge #100

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
Codeforces Round #665 (Div. 2) - A. Distance and Axis
https://codeforces.com/contest/1401/problem/A
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS=1e-9;
int INF=1000000005;
long long INFF=1000000000000000005ll;
double PI=acos(-1);
int dirx[8]={ -1, 0, 0, 1, -1, -1, 1, 1 };
int diry[8]={ 0, 1, -1, 0, -1, 1, -1, 1 };
const ll MOD = 1000000007;

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x) for(auto &(z):x) cin >> z;
#define FOR(a, b, c) for (int(a)=(b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a)=(b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a)=(b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a)=(b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a)=(b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto&(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a=max(a, b)
#define MIN(a, b) a=min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

void solve() {
ll n,k;
cin >> n >> k;
if(n<k) OUT(k-n);
// let m be the coordinate of B
// (n-m)-(m-0)=k
// m=(n-k)/2
// if both parity of n and k is even, then 0
// else 1
else OUT(!((n&1)^1==(k&1)^1));
// or just write
// else OUT(((n+k)&1));
}

int main()
{
FAST_INP;
//    #ifndef ONLINE_JUDGE
//    freopen("input.txt","r", stdin);
//    freopen("output.txt","w", stdout);
//    #endif

int tc; cin >> tc;
TC(tc) solve();

return 0;
}


/*
Codeforces Round #665 (Div. 2) - B. Ternary Sequence
https://codeforces.com/contest/1401/problem/B
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS=1e-9;
int INF=1000000005;
long long INFF=1000000000000000005ll;
double PI=acos(-1);
int dirx[8]={ -1, 0, 0, 1, -1, -1, 1, 1 };
int diry[8]={ 0, 1, -1, 0, -1, 1, -1, 1 };
const ll MOD = 1000000007;

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x) for(auto &(z):x) cin >> z;
#define FOR(a, b, c) for (int(a)=(b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a)=(b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a)=(b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a)=(b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a)=(b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto&(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a=max(a, b)
#define MIN(a, b) a=min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

void solve() {
vi a(3), b(3);
// c is either -2, 0, 2
// 2    for a=2; b=1
// -2   for a=1; b=2
// 0    else cases
// i.e. to make pair (1,0), (0,2) and (2,1) as much as possible
int ans=0;
int v = min(a[2],b[1]); // number of pair of (2,1) - produce 2 for each pair
a[2]-=v, b[1]-=v;
ans+= 2*v;

v = min(a[0],b[2]); // number of pair of (0,2) - produce 0 for each pair
a[0]-=v, b[2]-=v;

v = min(a[1],b[0]); // number of pair of (1,0) - produce 0 for each pair
a[1]-=v, b[0]-=v;

v = min(a[1],b[2]); // number of pair (1,2) - produce -2 for each pair
ans-= 2*v;
OUT(ans);
}

int main()
{
FAST_INP;
//    #ifndef ONLINE_JUDGE
//    freopen("input.txt","r", stdin);
//    freopen("output.txt","w", stdout);
//    #endif

int tc; cin >> tc;
TC(tc) solve();

return 0;
}


/*
Codeforces Round #665 (Div. 2) - C. Mere Array
https://codeforces.com/contest/1401/problem/C
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS=1e-9;
int INF=1000000005;
long long INFF=1000000000000000005ll;
double PI=acos(-1);
int dirx[8]={ -1, 0, 0, 1, -1, -1, 1, 1 };
int diry[8]={ 0, 1, -1, 0, -1, 1, -1, 1 };
const ll MOD = 1000000007;

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x) for(auto &(z):x) cin >> z;
#define FOR(a, b, c) for (int(a)=(b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a)=(b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a)=(b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a)=(b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a)=(b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto&(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a=max(a, b)
#define MIN(a, b) a=min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

void solve() {
int n; cin >> n;
vi a(n), b(n);
int mi = INF;
REP(i,n){
cin >> a[i];
b[i] = a[i];
MIN(mi, a[i]);
}
SORT(b);
int ok=1;
REP(i,n){
// if a[i] cannot be divided by the minimum element
// and it is out of order
// then it is NO
// else YES
if(a[i]%mi!=0&&a[i]!=b[i]){
ok=0;
break;
}
}
if(ok) OUT("YES");
else OUT("NO");
}

int main()
{
FAST_INP;
//    #ifndef ONLINE_JUDGE
//    freopen("input.txt","r", stdin);
//    freopen("output.txt","w", stdout);
//    #endif

int tc; cin >> tc;
TC(tc) solve();

return 0;
}


There are other programming solutions in the following repositories below. Star and watch for timely updates!

## wingkwong / timus

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### Wing-Kam

Consultant by day. Developer by night. AWS certified. Exploring #CloudNative currently.