Dragons | Editorial by k_m

Dragons | Editorial by k_m_madhav

**Problem Statement:**
The input consists of the character Kirito's initial strength, s and the number of dragons to defeat, n. The n subsequent lines consist of the dragon's strength, xi and the bonus strength received, yi.

**Approach:**
According to the problem statement, Kirito's strength must be *strictly* greater than the ith dragon's strength in order to defeat it. So the input xi and yi can be stored as a pair of integers and sorted in ascending order because it is logical to defeat the dragons of lower strength first in order to gain the bonus strength and proceed further. By default, the sorting is done such that the vector of pairs gets sorted based on the first item, i.e. the ith dragon's strength. We can further compare the first item from the vector of pairs and add the second term (which is the bonus strength) to initial variable, s and if true, increment a variable cnt. Finally, if the variable cnt equals n, the number of dragons to be defeated, then we output YES else NO.

**Code:**
```
#include<bits/stdc++.h>
#define fast ios::sync_with_stdio(false);cin.tie(NULL)
using namespace std;
#define int long long
#define MOD 1000000007
typedef vector<int> vi;
typedef pair<int,int> pi;
#define For(i,a,b) for (int i = a; i < b; i++)
#define Rep(i,a,b) for (int i = a; i <= b; i++)
#define ps(x,y) fixed << setprecision(y) << x 
#define pb push_back 
#define mp make_pair 
 
int32_t main()
{
    fast;
    int s, n, x[1005], y[1005], cnt = 0; cin >> s >> n;
    vector<pi> drags;
 
    For(i, 0, n) {
        cin >> x[i] >> y[i];
        drags.pb({x[i], y[i]});
        sort(drags.begin(), drags.end());
    }
 
    For(i, 0, n) {
        if(s > drags[i].first) {
            s += drags[i].second;
            ++cnt;
        }
    }
    
    cout << (cnt == n ? "YES" : "NO");
    
 
    return 0;
}
```

Dragons | Editorial by k_m_madhav

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