Codeforces Round ＃770 (Div. 2) ABC

Problem - A - Codeforces

The main idea of the topic ： Given a length of n String s, The number of transformations is k. ask k How many different strings can you get after one operation ？

Operating rules ：1. s→s+reverse（s）   or   2.  s→reverse（s）+ s ( Choose any transformation )

input

4
3 2
aab
3 3
aab
7 1
abacaba
2 0
ab

output

2
2
1
1

When the string is palindrome , Both transformations are ultimately the same , So just judge s Whether it is a palindrome string can determine the trend of future transformation . Non palindromes will also become palindromes after an operation , So it's one step more than palindrome .

Special judgement k==0 He still exists in a form of the original string

#include<iostream>
#include<cmath>
#include<string>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    int t; cin>>t;
    while(t--)
    {
        int n,k; cin>>n>>k;
        string s; cin>>s;
        bool flag=true;
        if(k==0) cout<<"1"<<endl;
        else{
                for(int i=0;i<n;i++)
                    if(s[i]!=s[n-i-1]) flag=false;
             if(flag==false) cout<<"2"<<endl;
             else cout<<"1"<<endl;
        }
    }
    return 0;
}

Problem - B - Codeforces

The main idea of the topic ： In limine Alice from x Start the operation ,Bob from x+3 Start the operation , And then there's a length of n Sequence a, Traverse this sequence from front to back , Every time Alice and Bob They have to make the following changes to their numbers ：

1.d→d+ai; 2.d→d xor ai（ A choice ）.

Ask after all the operations are finished , Who can just get y, And the output （ The title is guaranteed to be available to only one person y）.

input

4
1 7 9
2
2 0 2
1 3
4 0 1
1 2 3 4
2 1000000000 3000000000
1000000000 1000000000

output

Alice
Alice
Bob
Alice

adopt x And x+3 We can know that their parity is different , Each exclusive or operation on an odd number will change its parity , So we can assume alice Hold an odd number , And with arrays a Xor operation , If Alice After XOR with the whole array, and y The parity of is the same , Then output Alice , Otherwise output Bob.

Exclusive or ^ ：110  000  101  010

Bitwise AND & ：111 100 010 000

#include<iostream>
#include<cmath>
#include<string>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long LL;
LL a[200200];
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL t,n,x,y;
    cin>>t;
    while(t--)
    {
        cin>>n>>x>>y;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            x^=a[i];
        }
        if(((x^y)&1)==0)cout<<"Alice"<<endl;
        else cout<<"Bob"<<endl;
    }
    return 0;
}

Problem - C - Codeforces

The main idea of the topic ：n That's ok m Column , Ask if you can put 1~n*m Put them into this matrix and take out any string of numbers on any row , Their average must be an integer .

input

4
1 1
2 2
3 3
3 1

output

YES
1 
YES
1 3 
2 4 
NO
YES
1 
2 
3 

It can be simplified to two numbers ,/2 To be integer , That must be the same parity , Extended to a long string of numbers, it must be the same parity to get the whole . Then controlling the odd and even separation becomes .

#include<iostream>
#include<cmath>
#include<string>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    int t,n,m;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        if(m==1)
        {
            cout<<"YES"<<endl;
            for(int i=1;i<=n;i++)
                cout<<i<<endl;
        }
        else if(n&1) cout<<"NO"<<endl;// If it is an odd row, there must be a row with odd and even coexistence 
        else
        {
            cout<<"YES"<<endl;
            for(int i=1,k=1;i<=n*m;i+=2,k++)
            {
                cout<<i<<" ";
                if(k%m==0) cout<<endl;
            }
            for(int i=2,k=1;i<=n*m;i+=2,k++)
            {
                cout<<i<<" ";
                if(k%m==0) cout<<endl;
            }
        }
    }
    return 0;
}