AtCoder Beginner Contest 285解题报告

A - Edge Checker 2

Problem Statement

Determine if there is a segment that directly connects the points numbered a and b in the figure below.

Constraints

• 1≤a
• a and b are integers.

---

Input

The input is given from Standard Input in the following format:

a b

Output

Print Yes  if there is a segment that directly connects the points numbered a and b; print No otherwise.

---

Sample Input 1

1 2

Sample Output 1

Yes

In the figure in the Problem Statement, there is a segment that directly connects the points numbered 1 and 2, so Yes should be printed.

---

Sample Input 2 

2 8

Sample Output 2

No

In the figure in the Problem Statement, there is no segment that directly connects the points numbered 2 and 8, so No should be printed.

---

Sample Input 3

14 15

Sample Output 3

No

AC Code:

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
int a, b;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    cin >> a >> b;
    if (b == a * 2 || b == a * 2 + 1)
        puts("Yes");
    else
        puts("No");
    return 0;
}

B - Longest Uncommon Prefix

思路:朴素方法

AC Code:

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int n;
    string s;
    cin >> n >> s;
    for (int i = 1; i < n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (i + j > n)
            {
                cout << j - 1 << "\n";
                break;
            }
            if (s[j - 1] == s[j + i - 1])
            {
                cout << j - 1 << "\n";
                break;
            }
        }
    }
    return 0;
}

 C - abc285_brutmhyhiizp

 

 AC Code:

// 简单模拟
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    string s;
    cin >> s;
    int l = s.size();
    long long res = 0, add = 26;
    for (int i = 1; i <= l - 1; i++)
    {
        res += add;
        add *= 26;
    }
    long long num = 0;
    for (int i = 0; i < l; i++)
    {
        num *= 26;
        num += (s[i] - 'A');
    }
    cout << res + num + 1;
    return 0;
}

F - Substring of Sorted String

AC Code:

#include
using namespace std;

#define lowbit(x) x&(-x)

typedef long long ll;
const ll maxn=1e5+5;

int n,q;

char s[maxn];
int bit[30][maxn],sum[30],bit2[maxn];

void add(int a[maxn],int x,int c) {
	for(int i=x;i<=n;i+=lowbit(i)) a[i]+=c;
	return ;
}

int ask(int a[maxn],int x) {
	int res=0;
	for(int i=x;i>0;i-=lowbit(i)) res+=a[i];
	return res;
}

void add2(int x,int c) {
	for(int i=x;i0;i-=lowbit(i)) res+=bit2[i];
	return res;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	cin>>n>>s+1;
	for(int i=1;i<=n;i++) sum[s[i]-'a']++,add(bit[s[i]-'a'],i,1);
	for(int i=1;i>q;
	for(int i=1,op;i<=q;i++) {
		cin>>op;
		if(op==1) {
			int x;char c;
			cin>>x>>c;
			sum[s[x]-'a']--,add(bit[s[x]-'a'],x,-1);
			if(x1&&s[x-1]<=s[x]) add2(x-1,-1);
			s[x]=c;
			sum[s[x]-'a']++,add(bit[s[x]-'a'],x,1);
			if(x1&&s[x-1]<=s[x]) add2(x-1,1);
		}else {
			int l,r;
			cin>>l>>r;
			if(ask2(r-1)-ask2(l-1)!=r-l) {
				cout<<"No\n";
				continue;
			}
			int flag=0,tot=0;
			for(int j=0;j<26;j++) {
				int num=ask(bit[j],r)-ask(bit[j],l-1);
				tot+=num;
				if(!flag&&num==0) continue;
				if(!flag) flag=1;
				else {
					if(num