PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Author: Daanish Mahajan
Tester: Rahul Dugar
Editorialist: Aman Dwivedi
PROBLEM:
Chef is playing a T20 cricket match, and the opposite team has scored R runs in total. Chef has already scored C runs in O overs. There are a total of 20 overs that Chef can play, and in each over, there are 6 balls. The maximum runs that Chef can score in a single ball is 6 and assume that extras are not awarded in this match. Find if it is possible for the Chef to win the game if the winner has to score strictly higher runs than the opponent.
EXPLANATION:
In order to win the match, Chef should have a score strictly greater than the opponent’s score. The score of the opponent is already given to us as R. We are also given a score of C which Chef has scored in O overs.
Number of overs that are left in the game:
Chef can score 6 runs in each ball and hence 36 runs in an over. Hence the maximum runs that Chef can score in the remaining overs are:
Hence, the maximum score of Chef will be:
Now if the Chef score S is strictly greater than the opponent’s score R i.e (S>R), then Chef wins the game otherwise he loses.
TIME COMPLEXITY:
O(1)
SOLUTIONS:
Setter
#include<bits/stdc++.h>
using namespace std;
const int maxs = 720, maxo = 19;
int main()
{
int r, o, c; cin >> r >> o >> c;
assert(c <= 36 * o);
string ans = (r - c < 36 * (20 - o) ? "yEs" : "nO");
cout << ans << endl;
}
Tester
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#ifndef rd
#define trace(...)
#define endl '\n'
#endif
#define pb push_back
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef double f80;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<ll,ll> pll;
#define double long double
#define sz(x) ((long long)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
const int infi=0x3f3f3f3f;
const ll infl=0x3f3f3f3f3f3f3f3fLL;
const int mod=998244353;
//const int mod=1000000007;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
auto clk=clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0,lim-1);
return uid(rang);
}
int powm(int a, int b) {
int res=1;
while(b>0) {
if(b&1)
res=(res*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return res;
}
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 solve() {
int r=readIntSp(0,720),o=readIntSp(1,19),c=readIntLn(0,min(r,36*o));
string ans;
if(c+(20-o)*36>r) {
ans="yes";
} else {
ans="no";
}
for(char &i:ans) {
if(rng(2))
i=i-'a'+'A';
}
cout<<ans<<endl;
}
signed main() {
ios_base::sync_with_stdio(0),cin.tie(0);
srand(chrono::high_resolution_clock::now().time_since_epoch().count());
cout<<fixed<<setprecision(1);
// int t=readIntLn(1,100000);
int t=1;
// cin>>t;
fr(i,1,t)
solve();
assert(getchar()==EOF);
#ifdef rd
cerr<<endl<<endl<<endl<<"Time Elapsed: "<<((double)(clock()-clk))/CLOCKS_PER_SEC<<endl;
#endif
}
Editorialist
#include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int r,o,c;
cin>>r>>o>>c;
int rem=20-o;
if(c+rem*36>r)
cout<<"YES"<<"\n";
else
cout<<"NO"<<"\n";
}
int32_t main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
t=1;
while(t--)
solve();
return 0;
}