PROBLEM LINK:
Contest Division 1
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Contest Division 4
Setter: Utkarsh Gupta
Tester: Manan Grover, Abhinav sharma
Editorialist: Devendra Singh
DIFFICULTY:
1701
PREREQUISITES:
The average of N values is : \displaystyle\frac{Sum\:of\:all\:values}{N}
PROBLEM:
Chef has a set S containing N distinct integers.
Chef wants to gift Chefina an array A of any finite length such that the following conditions hold true:
- A_i \in S \forall i. In other words, each element of the array A should belong to the set S.
- Mean value of all the elements in A is exactly X.
Find whether there exists an array A of finite length satisfying the above conditions.
EXPLANATION:
Let min be the minimum of N given integers and max be the maximum of N given integers, then the average of these N integers always lies in the range [min, max]. Therefore if min>X or max< X the answer is NO. Otherwise the answer is always Yes.
Proof that the answer is always `yes` for the latter scenario:
Case 1: X is present in S, take a single occurrence of X, it has an average of X.
Case 2: X is not present in S,
Let a=X-min and b=max-X.
Then \displaystyle\frac{(a\cdot max + b\cdot min)}{(a + b)} = (\displaystyle\frac{(max-X)\cdot min+(X-min)\cdot max)}{(max-X+X-min)}=\displaystyle\frac{X\cdot (max-min)}{(max-min)} = X.
This means we can take a occurrences of the maximum element in S and b occurrences of the minimum element in S to get an average of X.
Thus, if min\leq X\leq max the answer is always Yes else No
TIME COMPLEXITY:
O(N) for each test case.
SOLUTION:
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,' ');
}
int sumN=0;
void solve()
{
int n=readInt(1,100000,' ');
int x=readInt(1,1000000000,'\n');
sumN+=n;
assert(sumN<=200000);
set <ll> s;
int maxi=0,mini=mod;
for(int i=1;i<=n;i++)
{
int c;
if(i==n)
c=readInt(1,1000000000,'\n');
else
c=readInt(1,1000000000,' ');
mini=min(mini,c);
maxi=max(maxi,c);
s.insert(c);
}
assert(s.size()==n);
if(x>=mini && x<=maxi)
cout<<"YES\n";
else
cout<<"NO\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');
//cin>>T;
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void sol(void)
{
int n, x, mi = 1e9, mx = 1;
cin >> n >> x;
vll v(n);
for (int i = 0; i < n; i++)
{
cin >> v[i];
mx = max(mx, v[i]);
mi = min(mi, v[i]);
}
cout<<((mx>=x && mi<=x)?"YES\n":"NO\n");
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}