Codeforces Round 951 (Div. 2)

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A - Guess the Maximum

直接暴力枚举 $a_i,a_{i+1}$ 找最小的最大值
答案即为最小的最大值-1

code:


#include<bits/stdc++.h>
#define endl '\n'
#define fast() ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define F first
#define S second
#define lowbit(x) ((x) & (-x))
#define _ int _T; cin >> _T; while(_T --) 
#define int long long
#define no cout << "NO" << endl
#define yes cout << "YES" << endlusing namespace std;
typedef pair<int, int> PII;
typedef pair<char,int> PCI;
const int N = 3e5 + 7;
const int mod = 998244353;
int n,m,k;
int a[N],b[N];
//PII p[N];void solve(){cin >> n;for(int i = 0;i < n;i ++) cin >> a[i];int mx = 1e18;for(int i = 0;i < n - 1;i ++){mx = min(mx,max(a[i],a[i + 1]));}cout << mx - 1 << endl;
}signed main(){fast();_solve();return 0;
}

B - XOR Sequences

序列长度无限所以从最低位到最高位按位枚举答案为小于最先不同的位数的所有值即 $(1 << i)$ 高位总会有 $\bigoplus$ 后使其相等的值

code:


#include<bits/stdc++.h>
#define endl '\n'
#define fast() ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define F first
#define S second
#define lowbit(x) ((x) & (-x))
#define _ int _T; cin >> _T; while(_T --) 
#define int long long
#define no cout << "NO" << endl
#define yes cout << "YES" << endlusing namespace std;
typedef pair<int, int> PII;
typedef pair<char,int> PCI;
const int N = 3e5 + 7;
const int mod = 998244353;
int n,m,k;
int a[N],b[N];
//PII p[N];void solve(){cin >> n >> m;for(int i = 0;i < 32;i ++){int cnt1 = (n >> i) & 1;int cnt2 = (m >> i) & 1;if(cnt1 != cnt2){cout << ((int)1 << i) << endl;return;}}  
}signed main(){fast();_solve();return 0;
}

C - Earning on Bets

假设每一位放的硬币数为 $a_i$ 那么总共就会放 $\sum\limits_{i = 1}^n{a_i}$ 个硬币
然后我们我们知道获胜会得到 $a_i*k_i$ 金币并且 $sum < a_i*k_i$
即 $\over k_i$ < $a_i$
并且有 $\sum\limits_{i = 1}^n{sum \over k_i} < \sum\limits_{i = 1}^n{a_i}$
即 $\sum\limits_{i = 1}^n{sum \over k_i} < sum$
那么有 $\sum\limits_{i = 1}^n{1 \over k_i} < 1$
那么我们可以令 $sum = lcm(k_1,k_2,k_3,...,k_n)$
这样 $sum * k_i$ 可以不剩余并且可以 $\over k_i$ 的分给每个位置的 $a_i$
即有 $\sum\limits_{i = 1}^n{sum \over k_i} < sum$ 满足此式子即说明有一个解
解为每个位置分 $\over k_i$

code:


#include<bits/stdc++.h>
#define endl '\n'
#define fast() ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define F first
#define S second
#define lowbit(x) ((x) & (-x))
#define _ int _T; cin >> _T; while(_T --) 
#define int long long
#define no cout << "NO" << endl
#define yes cout << "YES" << endlusing namespace std;
typedef pair<int, int> PII;
typedef pair<char,int> PCI;
const int N = 3e5 + 7;
const int mod = 998244353;
int n,m,k;
int a[N],b[N];
//PII p[N];void solve(){cin >> n;int lcm = 1;for(int i = 0;i < n;i ++){cin >> a[i];lcm = (a[i] * lcm) / __gcd(a[i],lcm);}  int sum = 0;for(int i = 0;i < n;i ++) sum += lcm / a[i];if(sum >= lcm){cout << -1 << endl;return;}for(int i = 0;i < n;i ++) cout << lcm / a[i] << " ";cout << endl;
}signed main(){fast();_solve();return 0;
}

D - Fixing a Binary String

直接从头开始遍历并且记录连续的0或1的个数
如果遍历过程中个数大于k了那么就有两种切割情况：

直接从刚大于时的位置前切开
看末尾的连续序列长度是否小于k 如果小于k并且 $s_i == s_n$ 那就说明此处可以拼接上去就从此序列开始位置+ (k - 末尾的连续序列长度 - 1)处切割

然后看这两种切割方式是否可以达成连续k个0或1交替出现即可
如果遍历过程中个数小于k了：

那么就在此前切割

code:


#include<bits/stdc++.h>
#define endl '\n'
#define fast() ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define F first
#define S second
#define lowbit(x) ((x) & (-x))
#define _ int _T; cin >> _T; while(_T --) 
#define int long long
#define no cout << "NO" << endl
#define yes cout << "YES" << endlusing namespace std;
typedef pair<int, int> PII;
typedef pair<char,int> PCI;
const int N = 3e5 + 7;
const int mod = 998244353;
int n,m,k;
int a[N],b[N];
//PII p[N];
int flag;void check(string ans){int len = 1;for(int i = 1;i < n;i ++){if(ans[i] == ans[i - 1]){len ++;if(len > k){flag = 1;break;}} else{if(len != k){flag = 1;break;}len = 1;}}    
}void solve(){cin >> n >> k;string s; cin >> s;int pos = -1,cnt = 1;int pos2 = -1;int st = 0;int mx = 1;for(int i = n - 2;i >= 0;i --){if(s[i] != s[i + 1]){break;} else{mx ++;}}for(int i = 1;i < n;i ++){if(s[i] == s[i - 1]){cnt ++;if(cnt > k){pos = i - 1;if(s[i] == s[n - 1]){if(mx >= k) pos2 = -1;else if(i - st >= k - mx) pos2 = st + k - mx - 1;}break;}} else{if(cnt != k){pos = i - 1;break;}st = i;cnt = 1;}}if(pos == -1){cout << n << endl;return;}string s1 = s.substr(0,pos + 1);string s2 = s.substr(pos + 1);reverse(s1.begin(),s1.end());string ans = s2 + s1;flag = 0;check(ans);if(!flag){cout << pos + 1 << endl;return;}if(pos2 == -1){cout << -1 << endl;return;}s1 = s.substr(0,pos2 + 1);s2 = s.substr(pos2 + 1);reverse(s1.begin(),s1.end());ans = s2 + s1;flag = 0;check(ans);if(!flag){cout << pos2 + 1 << endl;return;}cout << -1 << endl;
}signed main(){fast();_solve();return 0;
}