# Daily Coding Challenge #31

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 85 (Rated for Div. 2) - A. Level Statistics
https://codeforces.com/contest/1353/problem/A
*/

#include <bits/stdc++.h>
using namespace std;
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

int main()
{
FAST_INP;
int t,n,x,y;
cin >> t;
while(t--){
cin >> n;
int ans=1,px=0,py=0;
for(int i=0;i<n;i++){
cin >> x >> y;
if(
x<px || // the current number of plays cannot be smaller than the previous number of plays
y<py || // the current number of clears cannot be smaller than the previous number of clears
y-py > x-px  // number of clears cannot be smaller than the number of plays for one jump
) {
ans=0;
}

px=x; py=y;
}
if(ans) cout << "YES\n";
else cout << "NO\n";
}
return 0;
}

``````

``````/*
Educational Codeforces Round 85 (Rated for Div. 2) - B. Middle Class
https://codeforces.com/contest/1353/problem/B
*/

#include <bits/stdc++.h>
using namespace std;
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

int main()
{
FAST_INP;
int n,x,t;
cin >> t;
while(t--){
cin >> n >> x;
vector<int> a(n);
for(int i=0;i<n;i++) cin >>a[i];
// sort by descending order
sort(a.rbegin(),a.rend());
long long sum=0;
int idx=0;
for(int i=0;i<n;i++){
// if a[i] can be deducted, deduct it
if(a[i]>=x){
sum+=a[i]-x;
}else{
// if the largest number is smaller than x, set idx=-1
// if not, idx is the last idx of the number that can be deducted
idx=(i==0?-1:i);
break;
}
}

// the largest number is smaller than x
if(idx==-1){
cout << 0 << "\n";
continue;
}
// all numbers are greater than / equal to x
if(idx==0) {
cout << n << "\n";
continue;
}
int idx2=0;
// start from idx
for(int i=idx;i<n;i++){
// if the number can be added, add it up to x
if(sum>=x-a[i]) {
idx2++;
sum-=x-a[i];
}else{
break;
}
}
cout << idx+idx2 << "\n";
}
return 0;
}

``````

The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

### Discussion   