Codeforces Global Round 19

A. Sorting Parts

You have an array a of length n. You can exactly once select an integer len between 1 and n−1 inclusively, and then sort in non-decreasing order the prefix of the array of length len and the suffix of the array of length n−len

For example, if the array is a=[3,1,4,5,2], and you choose len=2, then after that the array will be equal to [1,3,2,4,5].

Could it be that after performing this operation, the array will not be sorted in non-decreasing order?
Input
There are several test cases in the input data. The first line contains a single integer t (1≤t≤100) — the number of test cases. This is followed by the test cases description.
The first line of each test case contains one integer n
(2≤n≤104) — the length of the array.

The second line of the test case contains a sequence of integers a1,a2,…,an (1≤ai≤109) — the array elements.

It is guaranteed that the sum of nover all test cases does not exceed 104.
Output
For each test case of input data, output “YES” (without quotes), if the array may be not sorted in non-decreasing order, output “NO” (without quotes) otherwise. You can output each letter in any case (uppercase or lowercase).
Example
Input

3
3
2 2 1
4
3 1 2 1
5
1 2 2 4 4

Output

YES
YES
NO

Note
In the first test case, it’s possible to select len=1, then after operation, the array will not be sorted in non-decreasing order and will be equal to [2,1,2].
In the second test case, it’s possible to select len=3, then after operation, the array will not be sorted in non-decreasing order and will be equal to [1,2,3,1].
In the third test case, the array will be sorted in non-decreasing order for every possible len.
思路：

大概就是判断这个数组是不是有序，有序就YES，无序就NO

#include <bits/stdc++.h>
using namespace std;
 
int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (auto& u : a)
            cin >> u;
        if (!is_sorted(a.begin(), a.end()))
            cout << "YES\n";
        else
            cout << "NO\n";
    }
}

B. MEX and Array

Let there be an array b1,b2,…,bk. Let there be a partition of this array into segments [l1;r1],[l2;r2],…,[lc;rc], where l1=1, rc=k, and for any 2≤i≤c holds that ri−1+1=li. In other words, each element of the array belongs to exactly one segment.
Let’s define the cost of a partition as c+∑i=1cmex({bli,bli+1,…,bri}),where mex of a set of numbers S is the smallest non-negative integer that does not occur in the set S. In other words, the cost of a partition is the number of segments plus the sum of MEX over all segments. Let’s define the value of an array b1,b2,…,bk as the maximum possible cost over all partitions of this array.You are given an array a of size n. Find the sum of values of all its subsegments.An array x is a subsegment of an array y if x can be obtained from y by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input

The input contains several test cases. The first line contains one integer t
(1≤t≤30) — the number of test cases.
The first line for each test case contains one integer n
(1≤n≤100) — the length of the array.
The second line contains a sequence of integers a1,a2,…,an
(0≤ai≤109) — the array elements.
It is guaranteed that the sum of the values n
over all test cases does not exceed 100.
Output

For each test case print a single integer — the answer to the problem.
Example
Input

4
2
1 2
3
2 0 1
4
2 0 5 1
5
0 1 1 0 1

Output

4
14
26
48

#include <bits/stdc++.h>
using namespace std;
 
int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (auto& u : a)
            cin >> u;
 
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans += (i + 1) * (n - i);
            if (a[i] == 0)
                ans += (i + 1) * (n - i);
        }
        cout << ans << '\n';
    }
}

C. Andrew and Stones

Andrew has n piles with stones. The i-th pile contains ai stones. He wants to make his table clean so he decided to put every stone either to the 1-st or the n-th pile.
Andrew can perform the following operation any number of times: choose 3 indices 1≤i<j<k≤n, such that the j-th pile contains at least 2 stones, then he takes 2 stones from the pile j and puts one stone into pile i and one stone into pile k.
Tell Andrew what is the minimum number of operations needed to move all the stones to piles 1and n, or determine if it’s impossible.
Input

The input contains several test cases. The first line contains one integer t
(1≤t≤10000) — the number of test cases.

The first line for each test case contains one integer n
(3≤n≤105) — the length of the array.

The second line contains a sequence of integers a1,a2,…,an
(1≤ai≤109) — the array elements.
It is guaranteed that the sum of the values n
over all test cases does not exceed 105.
Output

For each test case print the minimum number of operations needed to move stones to piles 1 and n, or print −1 if it’s impossible.

Input

4
5
1 2 2 3 6
3
1 3 1
3
1 2 1
4
3 1 1 2

Output

4
-1
1
-1 

#include <bits/stdc++.h>
using namespace std;
 
void solve() {
    
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &x : a)
        cin >> x;
 
    if (*max_element(a.begin() + 1, a.end() - 1) == 1 || (n == 3 && a[1] % 2 == 1)) {
    
        cout << "-1\n";
        return;
    }
 
    long long answer = 0;
    for (int i = 1; i < n - 1; i++)
        answer += (a[i] + 1) / 2;
 
    cout << answer << '\n';
}
 
int main() {
    
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int tests;
    cin >> tests;
    while (tests--)
        solve();
}