# Daily Coding Challenge #132

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 #672 (Div. 2) - A. Cubes Sorting
https://codeforces.com/contest/1420/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;

ll sum() { return 0; }
template<typename T, typename... Args>
T sum(T a, Args... args) { return a + sum(args...); }

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define OUTH(x) cout << x << " "
#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)
#define what_is(x) cerr << #x << " is " << x << endl;

void solve() {
int n, x;
cin >> n;
vi a(n);
int ok=0;
// to convert a[n] in a decreasing order to a non-decreasing order
// you need n(n-1)/2 operations
// ans is NO if the number of operations > n(n-1)/2 -1
// which means if there is a pair (i,j) where a[i] <= a[j]
// then the number of operations will be less than n(n-1)/2 -1
REP(i,n-1){
if(a[i]<=a[i+1]) {
ok=1;
break;
}
}
if(!ok) OUT("NO");
else OUT("YES");
}

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 #672 (Div. 2) - B. Rock and Lever
https://codeforces.com/contest/1420/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;

ll sum() { return 0; }
template<typename T, typename... Args>
T sum(T a, Args... args) { return a + sum(args...); }

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define OUTH(x) cout << x << " "
#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)
#define what_is(x) cerr << #x << " is " << x << endl;

void solve() {
int n;
cin >> n;
vi a(n);
ll ans=0;
unordered_map<int,int> m;
REP(i,n){
// if the largest set bit is same, then ai&aj >= ai^aj
// a 1000
// b 1010
//--------
// & 1000
// ^ 0010
// if the largest set bit does not match, then ai&aj < ai^aj
// a 1000
// b 0010
//--------
// & 0000
// ^ 1010
int k = (int)log2(a[i]);
// or
// int k = (x==0? -1 : 31 - __builtin_clz(a[i]));
ans+=m[k];
m[k]++;
}
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 #672 (Div. 2) - C1 - Pokemon Army (easy version)
https://codeforces.com/contest/1420/problem/C1
*/

#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;

ll sum() { return 0; }
template<typename T, typename... Args>
T sum(T a, Args... args) { return a + sum(args...); }

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define OUTH(x) cout << x << " "
#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)
#define what_is(x) cerr << #x << " is " << x << endl;

void solve() {
ll n, q;
cin >> n >> q;
vl a(n);
ll ans = a[0];
// observations:
// - optimal subsequence will be odd: + - +
// - get all increasing strengths
FOR(i,1,n) ans+=max(a[i]-a[i-1],0LL);
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;
}


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

## wingkwong / competitive-programming

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### Wing-Kam

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