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Practice
Setter: Utkarsh Gupta
Testers: Tejas Pandey and Abhinav sharma
Editorialist: Taranpreet Singh
DIFFICULTY
Simple
PREREQUISITES
Basic math
PROBLEM
Chef attempted an exam consisting of N objective questions. The marking scheme of the exam is:
- +3 marks for a correct answer.
- -1 marks for an incorrect answer.
- 0 marks for an unattempted question.
Find whether it is possible for Chef to score exactly X marks.
If it is possible, print 3 integers A, B, and C denoting the number of correct answers, incorrect answers, and unattempted questions respectively.
QUICK EXPLANATION
- We never need B \geq 3, since we can reduce A by 1, B by 3, and increase C by 4, leading to the same score.
- Considering each value of B in the range [0, 2], there’s exactly one value, which would make X+B a multiple of 3. Chef needs to answer (X+B)/3 questions correctly.
- Chef needs at least (X+B)/3 + B questions in order to achieve score X, so if (X+B)/3+B \gt N, then it is impossible.
- Lastly, we can choose C = N - ((X+B)/3+B) which will be non-negative.
EXPLANATION
Let’s try to write this down in form of equations. We need to find three integers A, B, and C s.t.
- A+B+C = N
- 3*A-B = X
- A, B, C \geq 0 (Number of questions cannot be negative).
- A, B, C are integers.
Since C is non-negative, we can convert first equation into inequality A+B \leq N.
Simpler problem
Consider following problem. Minimize A+B, subject to conditions
- A, B \geq 0
- 3*A-B = X
- A, B are integers.
An observation we can make is that if B \geq 3, then we can reduce A by 1 and B by 3, still satisfying 3*A-B = X and reducing A+B by 4.
Hence, it is never optimal to have B \geq 3. If there’s a solution with B \geq 3, there will be a solution with B \lt 3 as well.
So now, let’s try all values of B in the range [0, 2]. Since A = (X+B)/3 cannot be a fraction, only the value of B which makes X+B a multiple of 3 is viable. There would be exactly one such value in [0, 2]. Let’s call this p.
Hence, we have found B = p, and for that B, we can calculate A as (X+p)/3. So we have found (X+p)/3+p to be the minimum value of A+B.
Returning to the original question, A+B represents the sum of correct and incorrect answers. If we have (X+p)/3 + p \gt N, then Chef has attempted more than N questions, which is impossible. So no valid A, B and C exist in this case.
Otherwise, the number of attempted questions is (X+p)/3 + p, So we can assign C = N - ((X+p)/3 + p), making total number of questions equal to N, and score X, which is the required values.
Bonus
Prove that the resulting values will be A = \lceil \frac{X}{3} \rceil, B = 3*A-X and C = N-A-B. The answer would be impossible only when C \lt 0.
TIME COMPLEXITY
The time complexity is O(1) per test case.
SOLUTIONS
Setter's Solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
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;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
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 n=readInt(1,100000000,' ');
int x=readInt(0,3*n,'\n');
int correct=(x+2)/3;
int incorrect=(3*correct)-x;
if((correct+incorrect)>n)
cout<<"NO\n";
else
{
cout<<"YES\n";
cout<<correct<<' '<<incorrect<<' '<<(n-correct-incorrect)<<'\n';
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
int T=readInt(1,1000,'\n');
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Tester's Solution 1
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
/*
------------------------Input Checker----------------------------------
*/
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;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
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,' ');
}
/*
------------------------Main code starts here----------------------------------
*/
const int MAX_T = 1000;
const int MAX_N = 100000000;
#define ll long long int
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
long long int sum_len=0;
void solve()
{
int n = readIntSp(1, MAX_N);
int x = readIntLn(0, 3*n);
if((x + 2)/3 + ((3 - (x%3))%3) <= n) {
cout << "YeS\n";
cout << (x + 2)/3 << " " << ((3 - (x%3))%3) << " " << n - ((x + 2)/3 + ((3 - (x%3))%3)) << "\n";
} else {
cout << "nO\n";
}
}
signed main()
{
//fast;
#ifndef ONLINE_JUDGE
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
int t = readIntLn(1, MAX_T);
for(int i=1;i<=t;i++)
{
solve();
}
assert(getchar() == -1);
}
Tester's Solution 2
#include <bits/stdc++.h>
using namespace std;
/*
------------------------Input Checker----------------------------------
*/
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;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
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,' ');
}
/*
------------------------Main code starts here----------------------------------
*/
const int MAX_T = 1e5;
const int MAX_N = 1e5;
const int MAX_SUM_LEN = 1e5;
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ff first
#define ss second
#define mp make_pair
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define rev(i,n) for(int i=n;i>=0;i--)
#define rep_a(i,a,n) for(int i=a;i<n;i++)
int sum_n = 0, sum_m = 0;
int max_n = 0, max_m = 0;
int yess = 0;
int nos = 0;
int total_ops = 0;
ll mod = 998244353;
ll po(ll x, ll n){
ll ans=1;
while(n>0){ if(n&1) ans=(ans*x)%mod; x=(x*x)%mod; n/=2;}
return ans;
}
void solve()
{
int n = readIntSp(1, 1e8);
int x = readIntLn(0, 3*n);
int a,b,c;
a = (x+2)/3;
b = (x%3?(3-x%3):0);
if(a+b>n) cout<<"NO"<<'\n';
else{
cout<<"YES"<<'\n';
cout<<a<<" "<<b<<" "<<n-a-b<<'\n';
}
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r" , stdin);
freopen("output.txt", "w" , stdout);
#endif
fast;
int t = 1;
t = readIntLn(1,1000);
for(int i=1;i<=t;i++)
{
solve();
}
//assert(getchar() == -1);
assert(sum_n<=1e6);
cerr<<"SUCCESS\n";
cerr<<"Tests : " << t << '\n';
cerr<<"Sum of lengths : " << sum_n <<'\n';
cerr<<"Maximum length : " << max_n <<'\n';
// cerr<<"Total operations : " << total_ops << '\n';
//cerr<<"Answered yes : " << yess << '\n';
//cerr<<"Answered no : " << nos << '\n';
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class GENIUS{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int N = ni(), X = ni();
int A = (X+2)/3;
int B = A*3-X;
int C = (N-A-B);
if(A+B+C == N && C >= 0){
pn("YES");
pn(A+" "+B+" "+C);
}else pn("NO");
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
FastReader in;PrintWriter out;
void run() throws Exception{
in = new FastReader();
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new GENIUS().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
Feel free to share your approach. Suggestions are welcomed as always. 