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AtCoder 赛 236_atcoder航空路線図 (airline route map)

作者：盐析白兔 | 2024-02-09 17:33:43

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atcoder航空路線図 (airline route map)

A - chukodai

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : points100100

Problem Statement

You are given a string consisting of lowercase English letters.SS

Swap the -th and -th characters from the beginning of and print the resulting string.aabbSS

Constraints

SS is a string consisting of lowercase English letters.
The length of , , satisfies .SS|S|∣S∣2 \leq |S| \leq 102≤∣S∣≤10
1 \leq a < b \leq |S|1≤a<b≤∣S∣
aa and are integers.bb

Input

Input is given from Standard Input in the following format:

SS
aa bb

Output

Print the answer.

Sample Input 1

chokudai
3 5

Sample Output 1

chukodai

After swapping the -rd character and -th character of , we have .33o55uchokudaichukodai

Sample Input 2

aa
1 2

Sample Output 2

aa

In this sample, after swapping the -st and -nd characters of , we have the same string as .1122SSSS

Sample Input 3

aaaabbbb
1 8

Sample Output 3

baaabbba

思路:直接来,数组元素交换


#include<stdio.h>
int main(){
	char a[100];
	scanf("%s",a);
	int n,m;
	scanf("%d%d",&n,&m);
	n--;
	m--;
	char temp=a[n];
	a[n]=a[m];
	a[m]=temp;
	printf("%s",a);
}

可以用c++里的sawp,某大佬的源码


#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#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 pb push_back
#define fir first
#define sec second
#define SZ(x) ((int)x.size())
#define pii pair<int,int>
template<class T1,class T2>void ckmin(T1&x,T2 y){if(x>y)x=y;}
template<class T1,class T2>void ckmax(T1&x,T2 y){if(x<y)x=y;}
inline int read(){
    int x=0,f=0;char ch=getchar();
    while(!isdigit(ch))f|=ch=='-',ch=getchar();
    while(isdigit(ch))x=10*x+ch-'0',ch=getchar();
    return f?-x:x;
}
template<class T>void print(T x){
    if(x<0)putchar('-'),x=-x;
    if(x>=10)print(x/10);
    putchar(x%10+'0');
}
template<class T>void print(T x,char let){print(x),putchar(let);}
 
string s;int a,b;
 
int main(){
    cin>>s>>a>>b;
    a--,b--,swap(s[a],s[b]);
    cout<<s;
}

B - Who is missing?

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : points200200

Problem Statement

We have cards with an integer written on it, cards with , , cards with , for a total of cards.44114422\ldots…44NN4N4N

Takahashi shuffled these cards, removed one of them, and gave you a pile of the remaining cards. The -th card of the pile has an integer written on it.4N-14N−1ii(1 \leq i \leq 4N - 1)(1≤i≤4N−1)A_iAi

Find the integer written on the card removed by Takahashi.

Constraints

1 \leq N \leq 10^51≤N≤105
1 \leq A_i \leq N \, (1 \leq i \leq 4N - 1)1≤Ai≤N(1≤i≤4N−1)
For each , there are at most indices such that .k \, (1 \leq k \leq N)k(1≤k≤N)44iiA_i = kAi=k
All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN
A_1A1 A_2A2 \ldots… A_{4N - 1}A4N−1

Output

Print the answer.

Sample Input 1

3
1 3 2 3 3 2 2 1 1 1 2

Sample Output 1

Takahashi removed a card with written on it.33

Sample Input 2

1
1 1 1

Sample Output 2

Sample Input 3

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

Sample Output 3

思路:每种数有四个,直接用一个一个的桶装起来


#include<stdio.h>
int a[100001];
int main()
{
	int n;
	scanf("%d",&n);
	int k;
	for(int i=0;i<4*n-1;i++){
		scanf("%d",&k);
		a[k]++;
	}
	for(int i=1;i<=n;i++){
		if(a[i]!=4)
			printf("%d",i);
	}
}

c++


#include <bits/stdc++.h>
 
using namespace std;
 
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int x = 0;
  for (int i = 0; i < 4 * n - 1; i++) {
    int y;
    cin >> y;
    x ^= y;
  }
  cout << x << '\n';
  return 0;
}

使用x^=y,当y输入是两个相同的元素时,则x=0,所以可以统计出少的那个数(因为只会少一个数)

C - Route Map

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : points300300

Problem Statement

There are stations on a certain line operated by AtCoder Railway. The -th station from the starting station is named .NNii(1 \leq i \leq N)(1≤i≤N)S_iSi

Local trains stop at all stations, while express trains may not. Specifically, express trains stop at only stations, and the -th stop is the station named .
Here, it is guaranteed that and , that is, express trains stop at both starting and terminal stations.M \, (M \leq N)M(M≤N)jj(1 \leq j \leq M)(1≤j≤M)T_jTjT_1 = S_1T1=S1T_M = S_NTM=SN

For each of the stations, determine whether express trains stop at that station.NN

Constrains

2 \leq M \leq N \leq 10^52≤M≤N≤105
NN and are integers.MM
S_iSi (1 \leq i \leq N)(1≤i≤N) is a string of length between and (inclusive) consisting of lowercase English letters.111010
S_i \neq S_j \, (i \neq j)Si=Sj(i=j)
T_1 = S_1T1=S1 and .T_M = S_NTM=SN
(T_1, \dots, T_M)(T1,…,TM) is obtained by removing zero or more strings from and lining up the remaining strings without changing the order.(S_1, \dots, S_N)(S1,…,SN)

Input

Input is given from Standard Input in the following format:

NN MM
S_1S1 \ldots… S_NSN
T_1T1 \ldots… T_MTM

Output

Print lines. The -th line should contain if express trains stop at the -th station from the starting station, and otherwise.NNii(1 \leq i \leq N)(1≤i≤N)YesiiNo

Sample Input 1

5 3
tokyo kanda akiba okachi ueno
tokyo akiba ueno

Sample Output 1

Yes
No
Yes
No
Yes

Sample Input 2

7 7
a t c o d e r
a t c o d e r

Sample Output 2

Yes
Yes
Yes
Yes
Yes
Yes
Yes

Express trains may stop at all stations.

(时间超限一次).......

思路:火车是一直向前的所以出现过一次就不会再出现了


#include<stdio.h>
#include<string.h>
char a[100000][10];
char b[100000][10];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
		scanf("%s",a[i]);
	for(int i=0;i<m;i++)
		scanf("%s",b[i]);
	int i=0,j=0;
	while(j<m)
	{
		if(strcmp(a[i],b[j])==0){
			printf("Yes\n");
			j++;
		}
		else
			printf("No\n");
		i++;
	}
	return 0;
}

c++ 打表的方法,用map将数组名命名成字符串,map<string,int>mp;


#include <bits/stdc++.h>
 
using namespace std;
 
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<bool> res(n, false);
  map<string, int> mp;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    mp[s] = i;
  }
  for (int j = 0; j < m; j++) {
    string s;
    cin >> s;
    res[mp[s]] = true;
  }
  for (int i = 0; i < n; i++) {
    cout << (res[i] ? "Yes" : "No") << '\n';
  }
  return 0;
}

向量（Vector）是一个封装了动态大小数组的顺序容器（Sequence Container）。跟任意其它类型容器一样，它能够存放各种类型的对象。可以简单的认为，向量是一个能够存放任意类型的动态数组。

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