Codeforces Round #616 (Div. 1)

Mind Control

Topic links: https://codeforces.com/contest/1290/problem/A

Meaning of the questions:

　　A total of n items, each item has its own value.

　　N existing personal, everyone can be randomly or sequentially from scratch tail away a gift

　　M where you came in the first place, and you can specify that they take personal control X head or tail of gift

　　You can get at least ask how much value items

analysis:

　　We make K = min (X, M-1), it is because you can get a gift worth only by the personal influence of the former M-1

　　At this point you in front of you by K individuals can control, M - 1 - K individuals will not be out of your control, we make M - 1 - K = C

　　We let 0 <= i <= K, 0 <= j <= C, i mean you can control from the beginning and you let him take the gift of people Number

　　j means that you can not control the number of gifts and took him from the beginning, and then took the gift from the tail number is (K - i) + (C - j)

　　So to turn heads when you take the rest of the gift is a gift on a [i + j + 1], the rest of the tail as a gift to a [n - (K - i) - (Cj)]

　　Because the data is subject to a range of relatively small, so we can enumerate violence N ^ 2 friends

　　While enumerating the minimum we should pay attention because the number of j can not be controlled, so we have to take enumerate all the j

　　While the number i is that we can control, so we have to take the maximum of all i enumeration in

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 2e5 + 10;
ll a[N] , ans;
int main()
{
    int t , n , m , k , c , x;
    cin >> t;
    while(t --)
    {
        cin >> n >> m >> x ;
        for(int i = 1 ; i <= n ; i ++)    cin >> a[i];
        k = min(x , m - 1) , c = m - 1 - k , ans = 0 ;
        for(int i = 0 ; i <= k ; i ++)
        {
            ll now = (0x3f3f3f3f3f3fll);
            for(int j = 0 ; j <= c ; j ++)
                now = min(now , max(a[i + j + 1] , a[n - (k - i) - (c - j)]));
            ans = max(ans , now);
        }
        cout << ans << '\n';
    }
    return 0;
}

 

Irreducible Anagrams

Topic links: https://codeforces.com/contest/1290/problem/B

Meaning of the questions:　

　　Defined anagram, two characters refer to the same string.

　　Reducible defined anagram, represents two strings can be split into k sub-string and each substring is the anagram.

　　irreducible anagram that does not meet the criteria.

　　You are given a string, and then ask q times, each time asking for you l, r. Ask you [l, r] of this string, can find an irreducible anagram.

analysis:

　　Difficulty in item difficulty

　　Not difficult to find when the following is true irreducible anagram

　　①, a string of 1

　　②, the first string is not the same

　　③, there are three or more different character strings

　　①② direct determination operation O1, count the number of each character prefix pretreatment operation ③, each inquiry traversing it again

 

#include<bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<=n;i++)
using namespace std;
const int N = 3e5 + 10;
int cnt[N][37] , n , l , r;
string s;
int main()
{
    cin >> s;
    rep(i , 1 , s.size())
    {
        rep(j , 0 , 25)
        cnt[i][j] = cnt[i - 1][j];
        cnt[i][s[i - 1] - 'a'] ++ ;
    }
    cin >> n;
    while(n --)
    {
        cin >> l >> r;
        if(l == r) {cout << "Yes" << '\n' ; continue ;}
        if(s[l - 1] != s[r - 1]){cout << "Yes" << '\n' ; continue ;}
        int tot = 0;
        rep(j , 0 , 25)
        if(cnt[r][j] - cnt[l - 1][j])
        tot ++;
        if(tot >= 3) cout << "Yes" << '\n';
        else          cout << "No" << '\n';
    }
    return 0;
}

 

Summary: The first fight div1, the bottom of the T ^ T, points out much better