Codeworks round 650 (Div. 3) ABCD solution

Topic link ：

A. Short Substrings

B. Even Array

C. Social Distance

D. Task On The Board

A. Short Substrings

The question ： The string a All the lengths of are 2 The substring of is connected to form a new string , Calculate the original string a
Answer key ： Just skip the repeated

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
 
int t;
int main()
{
    
	cin>>t;
	while(t--){
    
		string a;
		cin>>a;
		cout<<a[0];
		for(int i=1;i<a.size();i++){
    
			cout<<a[i];
			if(i%2==1) i++;// Remove duplicate 
		}
		cout<<endl;
	}
	return 0;
 } 

B. Even Array

The question ： Swap element positions in the array , Make subscripts and elements parity , If you can output the number of exchanges , Otherwise output -1.
Answer key : Record the number of odd elements with even subscripts (q) And subscripts are even numbers with odd elements (w), If q=w, The output q, Anyway -1.

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int a;
int t,n;
int main()
{
    
cin>>t;
while(t--){
    
  cin>>n;
  int q=0,w=0;
  for(int i=0;i<n;i++){
    
  	cin>>a;
  	if(i%2==0&&a%2==1) q++;
	  else if(i%2==1&&a%2==0) w++; 
  }
  if(q==w) cout<<q<<endl;
  else if(q!=w)  cout<<-1<<endl;
 
}
 return 0;
 } 

C. Social Distance

The question ： Given a binary string , Make sure that each 1 Interval between k individual 0, Ask how many can be put 1.
Answer key ： Talk about four situations ：
1、 At present k+1 All for 0 when , Add... At the first position 1;
2、 If there is continuity in the middle 2k+1 individual 0 when , Add 1;
3、 If there is continuity at the end of the string k+1 And above individual 0 when , Add... At the end 1;
4、 If the string length is equal to k, The answer is 1;

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int a,x;
int t,n,k;
int main()
{
    
   cin>>t;
while(t--){
    
    cin>>n>>k;
    int q=0,w=0,c=0;
    char x;
    for(int i=0;i<n;i++){
    
    	cin>>x;
    	if(x=='0')  w++,q++;
    	else if(x=='1') q=0;
		if(q==k+1&&i==k) c++,q=k;
		// Case one 
		else if(q==2*k+1) c++,q=k;
		// The second case 
	}
	if(q>=k+1) c++;
	// The third case 
	if(w==n&&c==0) c++;
	// The fourth situation 
	cout<<c<<endl;
}
 return 0;
 } 

D. Task On The Board

The question ：For example, if t = “abzb”, then:
since t1=‘a’, all other indices contain letters which are later in the alphabet, that is: b1=|1−2|+|1−3|+|1−4|=1+2+3=6;
since t2=‘b’, only the index j=3 contains the letter, which is later in the alphabet, that is: b2=|2−3|=1;
since t3=‘z’, then there are no indexes j such that tj>ti, thus b3=0;
since t4=‘b’, only the index j=3 contains the letter, which is later in the alphabet, that is: b4=|4−3|=1.
Thus, if t = “abzb”, then b=[6,1,0,1].
Answer key ： Be careful b[j]=0 The point of , Because of 0, So he should put the maximum value .
We just need to put the largest letter in b[j]==0 It's about , Other places should be smaller than him , So the numbers in other places will become b[i]-=abs(i-j), And so on , Every time for 0 The points of all the remaining letters are the largest and the number is enough to be letters , Just keep cycling until the end .

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int q,m,x,b[60],v[27];
string a;
int vis[60];
char da[60];
int main()
{
    
	cin>>q;
	while(q--){
    
		memset(v,0,sizeof v);
		memset(vis,0,sizeof vis); 
		memset(da,'0',sizeof da);
		cin>>a>>m;
		for(int i=1;i<=m;i++) scanf("%d",&b[i]);
		for(int i=0;i<a.size();i++) v[a[i]-'a'+1]++; 
		// Subscripts indicate the size of letters , The value represents the number 
		int num=0,zero=0,k=27;
		while(num!=m){
    // Judge m Whether all positions have been placed 
			zero=0;
			int c=0;
			for(int i=1;i<=m;i++) if(b[i]==0&&vis[i]==0) zero++;
			// Record is currently 0 Number of 
			for(int i=k-1;i>0;i--) 
			if(v[i]>=zero) {
    k=i;break;}
			// Look for the largest and sufficient number of letters  
			for(int i=1;i<=m;i++) {
    
				if(b[i]==0&&vis[i]==0){
    
				vis[i]++;num++;
				// Record the location and number of storage 
				da[i]=k+'a'-1;
				// Store letters in 
					}
		}
		for(int i=1;i<=m;i++){
    
			if(b[i]==0&&da[i]==k+'a'-1) 
			for(int j=1;j<=m;j++) 
			if(b[j]) b[j]-=abs(j-i);
			// Other positions should be smaller than the currently added letters ,
			// So the position is poor 
		}
			}
		for(int i=1;i<=m;i++) cout<<da[i];
		cout<<endl;// Output 
	}
	return 0;
}