AtCoder 赛 236

AtCoder 赛 236

A - chukodai

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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∣≤101 leq a < b leq |S|1≤a<b≤∣S∣aa and are integers.bb

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Input

Input is given from Standard Input in the following format:

SS
aa bb

Output

Print the answer.

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Sample Input 1

chokudai
3 5

Sample Output 1

chukodai

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

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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

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Sample Input 3

aaaabbbb
1 8

Sample Output 3

baaabbba

思路:直接来,数组元素交换

#include
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
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
templatevoid ckmin(T1&x,T2 y){if(x>y)x=y;}
templatevoid ckmax(T1&x,T2 y){if(xvoid print(T x){
    if(x<0)putchar('-'),x=-x;
    if(x>=10)print(x/10);
    putchar(x%10+'0');
}
templatevoid 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< 
  
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.44114422ldots…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≤1051 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​=kAll 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 
    
3
 
    
Takahashi removed a card with written on it.33
 
    
   

    
    Sample Input 2 
    
1
1 1 1
 
    
    
    Sample Output 2 
    
1
 
    
   

    
    Sample Input 3 
    
4
3 2 1 1 2 4 4 4 4 3 1 3 2 1 3
 
    
    
    Sample Output 3 
    
2
 
    
思路:每种数有四个,直接用一个一个的桶装起来
 
    
#include
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 
 
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_jTj​T_1 = S_1T1​=S1​T_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≤105NN and are integers.MMS_iSi​ (1 leq i leq N)(1≤i≤N) is a string of length between and (inclusive) consisting of lowercase English letters.111010S_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
#include
char a[100000][10];
char b[100000][10];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i 
      
c++ 打表的方法,用map将数组名命名成字符串,mapmp;
 
      
#include 
 
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 res(n, false);
  map 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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