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Daily Coding Challenge #92

wingkwong profile image Wing-Kam ・4 min read

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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.


/*
Educational Codeforces Round 93 (Rated for Div. 2) - A. Bad Triangle
https://codeforces.com/contest/1398/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() {
    int n;
    cin >> n;
    vi a(n);
    READ(a);
    // sorted already
    // fix the first two and check the third one
    // UPD: we can just compare the last one 
    // i.e. a[0]+a[1]<=a[n-1]
    FOR(i,2,n){
        if(a[0]+a[1]<=a[i]){
            cout << 1 << " " << 2 << " " << (i+1) << endl;
            return;
        }
    }
    OUT(-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; 
} 

/*
Educational Codeforces Round 93 (Rated for Div. 2) - B. Substring Removal Game
https://codeforces.com/contest/1398/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() {
    string s;
    cin >> s;
    int cnt=0;
    vi score;
    // find the consecutive 1s and put them into a vector
    REP(i,s.size()){
        if(s[i]=='1') cnt++;
        else {
            score.pb(cnt);
            cnt=0;
        }
    }
    if(cnt) score.pb(cnt);
    // sort in descending order
    sort(score.rbegin(),score.rend());
    int ans=0;
    REP(i,score.size()){
        // add elements at even position
        if((i&1)^1) ans+=score[i];
    }
    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!

GitHub logo wingkwong / leetcode

πŸ† A Collection of my LeetCode Solutions with Explanations πŸ†

GitHub logo wingkwong / hackerrank

πŸ† A Collection of my HackerRank Solutions with Explanations πŸ†

GitHub logo wingkwong / codeforces

πŸ† A Collection of my Codeforces Solutions with Explanations πŸ†

GitHub logo wingkwong / atcoder

πŸ† A Collection of my AtCoder Solutions with Explanations πŸ†

GitHub logo wingkwong / cses

πŸ† A Collection of my CSES Solutions with Explanations πŸ†

GitHub logo wingkwong / timus

A Collection of my Timus Solutions

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wingkwong profile

Wing-Kam

@wingkwong

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

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