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Codeforces Round 163 (Rated for Div. 2)

A

按照题意，奇数肯定不行

偶数就不断输出AAB，每次有2个，满足题意

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const int N=1e6+100;
ll a[N];
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= ni--)
void solve()
{
int n;
cin>>n;
if(n&1)
{
    cout<<"NO"<<'\n';
}
else 
{
    int t=n/2;
    cout << "YES" << '\n';
    rep(i,1,t)
    {
        cout<<"AAB";
    }
cout<<'\n';
}
}
int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

B

感觉从尾枚举是最优的

因为尾部大就不需要分

尾部比前面一个小

一定要分

因此从尾部开始枚举

如果前面的数比尾数大并且不满足分的性质（个位数大于等于十位）或者分完依然大，那么就没救了

如果可以，那么把当前的数字赋值为十位数

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n;i--)
const int N=1e4+100;
struct 
{
    int v;
    int l,r;
    bool vis;
}a[N];
void solve()
{
int n;
cin>>n;
rep(i,1,n)
{
    cin>>a[i].v;
    if(a[i].v>=10)
    {
        a[i].l=a[i].v/10;
        a[i].r=a[i].v%10;
if(a[i].l>a[i].r)
{
a[i].vis=0;
}
else
{
    a[i].vis=1;
}
    }
}
frep(i,n,2)
{
if(a[i-1].v>a[i].v)
{
if(a[i-1].vis&&a[i-1].r<=a[i].v)
{
    a[i-1].v=a[i-1].l;
}
else
{
    cout<<"NO"<<'\n';
    return ;
}
}
}
cout<<"YES"<<'\n';
}
int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

C

C题的话实际上感觉和之前某场很像

数据挺大

赛后坤佬说可bfs

实际上我不是用bfs做的

观察奇数点才是我们要走的

记录下来

如果旁边有两0就不行

加点特判

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= ni--)
const int N = 1e6;
char a[3][N];
int b[3][N];
void solve()
{
    int n;
    cin >> n;
    rep(i, 1, 2)
    {
        rep(j, 1, n)
    {
        cin >> a[i][j];
    }
    }
    n /= 2;
    rep(i, 1, 2)
    {
        rep(j, 1, n)
        {
            if (i == 1)
            {
                if (a[i][j * 2] == '>')
                {
                    b[i][j] = 1;
                }
                else
                {
                    b[i][j] = 0;
                }
            }
        else
        {
            if (a[i][j * 2 - 1] == '>')
            {
                b[i][j] = 1;
            }
            else
            {
                b[i][j] = 0;
            }
        }
    }
}
if (b[1][n] == 0 && b[2][n] == 0)
{
    cout << "NO" << '\n';
    return ;
}
if(b[1][1]==0&&b[2][1]==0)
{
    cout<<"NO"<<'\n';
    return ;
}
rep(i,1,2)
{
    rep(j,1,n)
   {
   if(i==1)
   {
    if(j==1)
    {
        continue;
    }
if(b[i][j-1]==0&&b[i+1][j]==0)
{
  cout<<"NO"<<'\n';
  return ;  
}
   }
else if(i==2)
{
    if(j==1)
    {
        continue;
    }
if(b[i-1][j-1]==0&&b[i][j-1]==0)
{
    cout<<"NO"<<'\n';
    return ;
}
}
    }
   }
cout<<"YES"<<'\n';
}
int main()
{
std::ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while (t--)
{
    solve();
    }
}

D

D题暴力枚举相等长度即可

？只是多了个判断罢了

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n;i--)
const int N = 1e6 + 10;
void solve()
{
    string s;
    cin >> s;
    int ans = 0;
    frep(i,s.length()/2,1)
    {
        int num = 0;
        for (int j = 0; j + i < s.length(); j++)
        {
            if (s[j] == s[j + i] || s[j] == '?' || s[j + i] == '?')
            {
                num++;
                if (num == i)
                {
                    ans = max(ans, i);
                }
            }
            else
            {
                num = 0;
            }
        }
    }
    cout << ans * 2 << "\n";
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}