Codeforces Round #590 (Div. 3) codeforces打完补题

https://codeforces.com/contest/1234/problem/D

写了个巨蠢的线段树（不愧是垃圾），有必要提醒下自己这种题怎么做

#include
#include
#include
#include
#include
using namespace std;
const int maxn = 32;
set st[maxn];
int main()
{
    string words;
    cin>>words;
    words = '.' + words;
    int len = words.size() - 1;
    for(int i=1;i<=len;++i)
        st[words[i]-'a'].insert(i);
    int q,cmd,pos,l,r;
    char ch;  scanf("%d",&q);
    for(int k=0;k!=q;++k)
    {
        scanf("%d",&cmd);
        if(cmd==1)
        {
            scanf("%d",&pos);
            getchar();
            scanf("%c",&ch);
            st[words[pos]-'a'].erase(pos);
            words[pos] = ch;
            st[words[pos]-'a'].insert(pos);
        }else{
            int ans = 0;
            scanf("%d%d",&l,&r);
            for(int i=0;i!=26;++i)
            {
                auto it = st[i].lower_bound(l);//MDZZ 再也不直接用algorithm的二分了,容器内的快多了,佛了
                if(it == st[i].end())
                    continue;//找到大于等于l开始的字符,因为不重复,所以只需要找出一个就好了(ORZ)
                int x = *it;
                if(x <= r)
                    ++ans;
            }
            cout< 
 

https://codeforces.com/contest/1234/problem/C

　简单实现

#include
using namespace std;
const int maxn = 2e5 + 32;
int dp[3][maxn];
 int q,n;
void solve()
{
    int l = 1,r = 1,k = 1;
    while(1)
    {
        if(l>n)
        {
            if(r==2)
            {
                cout<<"YES"<<'\n';
                return;
            }
            cout<<"NO"<<'\n';
            return;
        }
        if(k==1&&dp[r][l]==1)
        {++l;}else 
        if(k==1&&dp[r][l]==2)
        {r = 3 - r; k = 2;}else
        {
            if(dp[r][l]!=2)
            {
                cout<<"NO"<<'\n';
                return;
            }else{
                ++l;
                k = 1;
            }
        } 
    }
}
int main()
{
    ios::sync_with_stdio(false); cin.tie(0),cout.tie(0);
    cin >> q;
    while(q--)
    {
        cin >> n;   char ch;
        for(int i=1;i<=2;++i)
        {
            for(int j=1;j<=n;++j)
            {
                cin>>ch;
                if(ch=='1'||ch=='2')
                    dp[i][j] = 1;
                else
                    dp[i][j] = 2;
            }
        }
        solve();
    }
}

　　

　垃圾如我，哭啦