PROBLEM LINK:
Contest Division 1
Contest Division 2
Contest Division 3
Setter: Jeevan Jyot Singh
Tester: Aryan Choudhary
Editorialist: Lavish Gupta
DIFFICULTY:
Cakewalk
PREREQUISITES:
None
PROBLEM:
There are 101 citizens in Chefland. It is election time in Chefland and 3 parties, A, B, and C are contesting the elections. Party A receives X_A votes, party B receives X_B votes, and party C receives X_C votes.
The constitution of Chefland requires a particular party to receive a clear majority to form the government. A party is said to have a clear majority if it receives strictly greater than 50 votes.
If any party has a clear majority, print the winning party (A
, B
or C
). Otherwise, print NOTA
.
EXPLANATION:
The aim of the problem is to test the ability of converting the problem statement into an error-free running code. Note the following observation:
- There can be at max 1 party which have a clear majority. More than 1 party cannot have clear majority.
So we can check if X_A > 50 using if condition.
If Yes, the answer is A.
If No, then check for B, and then C if required.
If none of X_A, X_B, X_C are strictly greater than 50, then the answer is NOTA
.
Tip for beginners
Avoid using nested if-else conditions. Try to use simple if conditions to solve the problem.
Corner Test Case
1
50 49 2
The answer should be NOTA
. The key-point is, we need X_i > 50, and not X_i \geq 50.
TIME COMPLEXITY:
Each Test-Case will require checking constant number of conditions. Therefore it will take O(1) time for each test case.
SOLUTION:
Tester's Solution
/* in the name of Anton */
/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/
#ifdef ARYANC403
#include <header.h>
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif
// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
Fun fun_;
public:
template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }
using namespace std;
#define fo(i,n) for(i=0;i<(n);++i)
#define repA(i,j,n) for(i=(j);i<=(n);++i)
#define repD(i,j,n) for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"
typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;
const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}
long long readInt(long long l, long long r, char endd) {
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
return readInt(l,r,' ');
}
long long readIntLn(long long l, long long r) {
return readInt(l,r,'\n');
}
string readStringLn(int l, int r) {
return readString(l,r,'\n');
}
string readStringSp(int l, int r) {
return readString(l,r,' ');
}
void readEOF(){
assert(getchar()==EOF);
}
vi readVectorInt(int n,lli l,lli r){
vi a(n);
for(int i=0;i<n-1;++i)
a[i]=readIntSp(l,r);
a[n-1]=readIntLn(l,r);
return a;
}
// #include<atcoder/dsu>
// vector<vi> readTree(const int n){
// vector<vi> e(n);
// atcoder::dsu d(n);
// for(lli i=1;i<n;++i){
// const lli u=readIntSp(1,n)-1;
// const lli v=readIntLn(1,n)-1;
// e[u].pb(v);
// e[v].pb(u);
// d.merge(u,v);
// }
// assert(d.size(0)==n);
// return e;
// }
const lli INF = 0xFFFFFFFFFFFFFFFL;
lli seed;
mt19937 rng(seed=chrono::steady_clock::now().time_since_epoch().count());
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}
class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{ return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y )); }};
void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end()) m.insert({x,cnt});
else jt->Y+=cnt;
}
void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt) m.erase(jt);
else jt->Y-=cnt;
}
bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}
const lli mod = 1000000007L;
// const lli maxN = 1000000007L;
lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
lli m;
string s;
vi a;
//priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .
int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
T=readIntLn(1,500);
const lli sumVotes = 101;
while(T--)
{
const lli A=readIntSp(0,sumVotes);
const lli B=readIntSp(0,sumVotes-A);
const lli C=sumVotes-A-B;
readIntLn(C,C);
if(2*A>sumVotes){
cout<<"a"<<endl;
continue;
}
if(2*B>sumVotes){
cout<<"b"<<endl;
continue;
}
if(2*C>sumVotes){
cout<<"c"<<endl;
continue;
}
cout<<"NoTa"<<endl;
} aryanc403();
readEOF();
return 0;
}
Editorialist's Solution
#include<bits/stdc++.h>
using namespace std ;
int main()
{
int t ;
cin >> t ;
while(t--)
{
int a , b , c ;
cin >> a >> b >> c ;
if(a > 50)
{
cout << "A\n" ;
continue ;
}
if(b > 50)
{
cout << "B\n" ;
continue ;
}
if(c > 50)
{
cout << "C\n" ;
continue ;
}
cout << "NOTA\n" ;
}
return 0;
}