AtCoder Beginner Contest 240

比赛链接
在这强烈dls的视频讲解
dls的视频讲解及解题过程

A - Edge Checker

Problem Statement
In the figure shown in the image below, are the points numbered
a and b directly connected by a line segment?
Input
Input is given from Standard Input in the following format:
a b
Output

If the points numbered
a and b are directly connected by a line segment, print Yes; otherwise, print No.
The judge is case-sensitive: be sure to print uppercase and lowercase letters correctly.

思路
看两个数字是否在同一条线上。

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define PII pair<int, int>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define rep2(i, l, r) for (int i = l; i * i <= r; i++)
#define rep3(i, l, r) for (LL i = l; i * i * i <= r; i++)
#define Min(a, b) a > b ? b : a
#define Max(a, b) a > b ? a : b
#define endl '\n'
#define debug "-----"
using namespace std;
typedef long long LL;
const LL mod = 1e9;
const int N = 1e6 + 10, M = 1050;
LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }
int main()
{
	IOS;
	int a,b;
	cin >> a >> b;
	if( abs(a-b)==1 || ( a==10&&b==1 ) || (b==10&&a==1) )
		cout << "Yes";
	else cout << "No";
	return 0;
}

B - Count Distinct Integers

思路：
找不同的数字有多少个

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define PII pair<int, int>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define rep2(i, l, r) for (int i = l; i * i <= r; i++)
#define rep3(i, l, r) for (LL i = l; i * i * i <= r; i++)
#define Min(a, b) a > b ? b : a
#define Max(a, b) a > b ? a : b
#define endl '\n'
#define debug "-----"
using namespace std;
typedef long long LL;
const LL mod = 1e9;
const int N = 1e6 + 10, M = 1050;
LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }

map<int,int>mp;
int ans = 0;

int main()
{
	IOS;
	int n;
	cin >> n;
	rep( i , 1 , n+1 ){
		int t; cin >> t;
		mp[t]++;
		if( mp[t] == 1 ) ans++;
	}
	cout << ans;
	return 0;
}

C - Jumping Takahashi

思路
当初自己写这道题的时候用了DFS结果超时了。一直想到了结束，自己还是太菜了QAQ。
在这推荐下dls Atcoder的视频讲解
dls的题解

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define PII pair<int, int>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define rep2(i, l, r) for (int i = l; i * i <= r; i++)
#define rep3(i, l, r) for (LL i = l; i * i * i <= r; i++)
#define Min(a, b) a > b ? b : a
#define Max(a, b) a > b ? a : b
#define endl '\n'
#define debug "-----"
using namespace std;
typedef long long LL;
const LL mod = 1e9;
const int N = 1e6 + 10, M = 1050;
LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }

bitset<10001>f;

int main()
{
	IOS;
	f[0] = 1;
	int n,x;
	cin >> n >> x;
	rep( i , 1 , n+1){
		int a,b;
		cin >> a >> b;
		f = ( f<<a )|( f<<b );
	}
	f[x]?cout << "Yes":cout << "No";
	return 0;
}

D - Strange Balls

思路：
模拟及一些优化，感觉有点类似kmp的next

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define PII pair<int, int>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define rep2(i, l, r) for (int i = l; i * i <= r; i++)
#define rep3(i, l, r) for (LL i = l; i * i * i <= r; i++)
#define Min(a, b) a > b ? b : a
#define Max(a, b) a > b ? a : b
#define endl '\n'
#define debug "-----"
using namespace std;
typedef long long LL;
const LL mod = 1e9;
const int N = 1e6 + 10, M = 1050;
LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }

int s[N],stk[N];
int hh;

int main()
{
	IOS;
	int n;
	cin >> n;
	rep( i , 1 , n+1 ){
		int t;
		cin >> t;
		stk[++hh] = t;
		if( hh == 1 || stk[hh]!=stk[hh-1] ) s[hh] = 1;
		else s[hh] = s[hh-1]+1;

		if( s[hh] == t ) hh -= t;
		cout << hh << endl;
	}
	return 0;
}