【codeforces 479DIV.3】A B C E题解

Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:

if the last digit of the number is non-zero, she decreases the number by one;
if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).

You are given an integer number $n$ . Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.

It is guaranteed that the result will be positive integer number.

Input

The first line of the input contains two integer numbers $n$ and $k$ ( $2 \leq n \leq 10^{9}$ , $1 \leq k \leq 50$ ) — the number from which Tanya will subtract and the number of subtractions correspondingly.

Output

Print one integer number — the result of the decreasing $n$ by one $k$ times.

It is guaranteed that the result will be positive integer number.

      Examples 
    
        input 
       
         Copy 
       
512 4

        output 
       
         Copy 
       
50

        input 
       
         Copy 
       
1000000000 9

        output 
       
         Copy 
       
1

Note

The first example corresponds to the following sequence: $512 \to 511 \to 510 \to 51 \to 50$

简单的模拟即可，如果n能被10整除，则n除10，否则n-1；

#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
using namespace std;
int main()
{
	int n,t;
	cin>>n>>t;
	while(t--)
	{
		if(n%10==0)
		{
			n/=10;
		}
		else
		{
			n--;
		}
	}
	cout<<n<<endl;
	return 0;
}

Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.

You are given a string $s$ consisting of $n$ capital Latin letters. Your task is to find any two-gram contained in the given string as a substring(i.e. two consecutive characters of the string) maximal number of times. For example, for string $s$ = "BBAABBBA" the answer is two-gram "BB", which contained in $s$ three times. In other words, find any most frequent two-gram.

Note that occurrences of the two-gram can overlap with each other.

Input

The first line of the input contains integer number $n$ ( $2 \leq n \leq 100$ ) — the length of string $s$ . The second line of the input contains the string $s$ consisting of $n$ capital Latin letters.

Output

Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string $s$ as a substring (i.e. two consecutive characters of the string) maximal number of times.

       Examples 
     
         input 
        
          Copy 
        
7
ABACABA

         output 
        
          Copy 
        
AB

         input 
        
          Copy 
        
5
ZZZAA

         output 
        
          Copy 
        
ZZ

Note

In the first example "BA" is also valid answer.

In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.

直接暴力跑一遍，用map记录两个字符形成的串的个数,找出最大值即可：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
#include<map>
using namespace std;
char c[105];
map<string,int> p;
int main()
{
		int n;
		cin>>n;
		for(int i=0;i<n;i++)
		{
			cin>>c[i];
		}
		int maxx=0;
		char re[2],ss[2];
		for(int i=0;i<n-1;i++)
		{
			ss[0]=c[i];
			ss[1]=c[i+1];
			p[ss]++;
			if(p[ss]>maxx)
			{
				maxx=p[ss];
				strcpy(re,ss);
			}
		}
		for(int i=0;i<2;i++)
		{
			cout<<re[i];
		}	
	return 0;
 }

You are given a sequence of integers of length $n$ and integer number $k$ . You should print any integer number $x$ in the range of $[1; 10^{9}]$ (i.e. $1 \leq x \leq 10^{9}$ ) such that exactly $k$ elements of given sequence are less than or equal to $x$ .

Note that the sequence can contain equal elements.

If there is no such $x$ , print "-1" (without quotes).

Input

The first line of the input contains integer numbers $n$ and $k$ ( $1 \leq n \leq 2 \cdot 10^{5}$ , $0 \leq k \leq n$ ). The second line of the input contains $n$ integer numbers $a_{1}, a_{2}, \dots, a_{n}$ ( $1 \leq a_{i} \leq 10^{9}$ ) — the sequence itself.

Output

Print any integer number $x$ from range $[1; 10^{9}]$ such that exactly $k$ elements of given sequence is less or equal to $x$ .

If there is no such $x$ , print "-1" (without quotes).

    Examples 
  
      input 
     
       Copy 
     
7 4
3 7 5 1 10 3 20

      output 
     
       Copy 
     
6

      input 
     
       Copy 
     
7 2
3 7 5 1 10 3 20

      output 
     
       Copy 
     
-1

Note

In the first example $5$ is also a valid answer because the elements with indices $[1, 3, 4, 6]$ is less than or equal to $5$ and obviously less than or equal to $6$ .

In the second example you cannot choose any number that only $2$ elements of the given sequence will be less than or equal to this number because $3$ elements of the given sequence will be also less than or equal to this number.

注意几种特殊情况的判断，直接排序即可，判断a[k-1]是否等于a[k]，判断k=0和k=n的情况即可

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int a[200005];
int main()
{
	int n,k;
	cin>>n>>k;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	sort(a,a+n);
	if(k==0) 
	{
		if(a[0]<=1) cout<<"-1"<<endl;
		else cout<<"1"<<endl;
	}
	else if(k==n)
	{
		cout<<"1000000000"<<endl;
	}
	else
	{
		if(a[k-1]==a[k]) cout<<"-1"<<endl;
		else 
		{
			cout<<a[k-1]<<endl;
		}
	}
	return 0;
}

You are given an undirected graph consisting of $n$ vertices and $m$ edges. Your task is to find the number of connected components which are cycles.

Here are some definitions of graph theory.

An undirected graph consists of two sets: set of nodes (called vertices) and set of edges. Each edge connects a pair of vertices. All edges are bidirectional (i.e. if a vertex $a$ is connected with a vertex $b$ , a vertex $b$ is also connected with a vertex $a$ ). An edge can't connect vertex with itself, there is at most one edge between a pair of vertices.

Two vertices $u$ and $v$ belong to the same connected component if and only if there is at least one path along edges connecting $u$ and $v$ .

A connected component is a cycle if and only if its vertices can be reordered in such a way that:

the first vertex is connected with the second vertex by an edge,
the second vertex is connected with the third vertex by an edge,
...
the last vertex is connected with the first vertex by an edge,
all the described edges of a cycle are distinct.

A cycle doesn't contain any other edges except described above. By definition any cycle contains three or more vertices.

$\text{[math]}$ There are

6

connected components,

2

of them are cycles:

[7, 10, 16]

and

[5, 11, 9, 15]

Input

The first line contains two integer numbers $n$ and $m$ ( $1 \leq n \leq 2 \cdot 10^{5}$ , $0 \leq m \leq 2 \cdot 10^{5}$ ) — number of vertices and edges.

The following $m$ lines contains edges: edge $i$ is given as a pair of vertices $v_{i}$ , $u_{i}$ ( $1 \leq v_{i}, u_{i} \leq n$ , $u_{i} \neq v_{i}$ ). There is no multiple edges in the given graph, i.e. for each pair ( $v_{i}, u_{i}$ ) there no other pairs ( $v_{i}, u_{i}$ ) and ( $u_{i}, v_{i}$ ) in the list of edges.

Output

Print one integer — the number of connected components which are also cycles.

Note

In the first example only component $[3, 4, 5]$ is also a cycle.

The illustration above corresponds to the second example.

利用并查集查找环的个数，不过要注意环是独环，所以每个点的度要为2才行，所以这还是一个坑点。

#include <stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std; 
int fa[200005];
int find(int x)
{
	if(fa[x]!=x) fa[x]=find(fa[x]);
	return fa[x];
}
bool comb(int a,int b)
{
	a=find(a);
	b=find(b);
	if(a==b)//如果根相等就是出现了环
	return true;
	else
	{
		fa[a]=b;
		return false;
	}
}
void init(int n)
{
	for(int i=0;i<n;i++)
	fa[i]=i;
}
int du[200005];
int u[200005],v[200005];
int main()
{
	int n,k;
	while(cin>>n>>k)
	{
		memset(du,0,sizeof(du));
		init(n);
		int sum=0;
		for(int i=0;i<k;i++)
		{
			cin>>u[i]>>v[i];
			du[u[i]]++;
			du[v[i]]++;
		}
		for(int i=0;i<k;i++){
			if(du[u[i]]==2&&du[v[i]]==2&&(comb(u[i],v[i])))
			{
				sum++;
			}
		}
		cout<<sum<<endl;
	}
	return 0;
}

【codeforces 479DIV.3】A B C E题解

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