G.The Witchwood

思路：

從大到小排序，把前k個數加起來

#include <bits/stdc++.h>

using namespace std;

const int N = 1010;

int a[N];

int main()
{
    int n, k;
    cin >> n >> k;
    for(int i = 0; i < n; i ++ ) cin >> a[i];
    sort(a, a + n, greater<int>());
    
    long long sum = 0;
    
 
for(int i = 0; i < k; i ++ ) sum += a[i];
    cout << sum << endl;
    
    return 0;
}

F.Kobolds and Catacombs

思路：

記錄一下原陣列和排好序後的陣列，並記錄其字首和，每次字首和相等的時候就答案數加一

#include <bits/stdc++.h>

using namespace std;

const int N = 1000010;

int a[N], b[N];

int main()
{
    int n;
    cin >> n;
    
 
for(int i = 0; i < n; i ++ ) cin >> a[i], b[i] = a[i];
    sort(a, a + n);
    long long suma = 0, sumb = 0, res = 0;
    for(int i = 0; i < n; i ++ )
    {
        suma += a[i];
        sumb += b[i];
        if(suma == sumb) res ++ ;
    }
    cout << res << endl;
    
    return
 
 0;
}