多校A层冲刺NOIP2024模拟赛04

多校A层冲刺NOIP2024模拟赛04

学校OJ赛时 rank28，T1 0pts，T2 100pts，T3 20pts,T4 25pts
accoders rank15，T1 100pts，T2 100pts，T3 20pts，T4 25pts

不是，也没人告诉我两个OJ文件名不一样啊

02表示法

递归 + 高精除低精。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = infile("power.in"),*OutFile = outfile("power.out");
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 200;
string s;
ll n;
inline int divide(string &s){
    if(!s.size()) return -1;
    if(s.size() == 1 && s[0] == '0') return -1;
    if(s.size() == 1 && s[0] == '1'){
        s = "";return 1;
    }
    int res = (s.back() - '0')%2;s.back() -= res;
    int n = s.size(),x = 0;
    string sh;bool flag = false;
    rep(i,0,n-1,1){
        x = x*10 + (s[i] - '0');
        if(x >= 2) sh.push_back(x/2+'0'),x %= 2,flag = true;
        else if(flag) sh.push_back('0');
    }
    s = sh;
    return res;
}
#define pii pair<int,int>
#define mk make_pair
void work(string s){
    if(s == "0") return cout<<0,void();
    if(s == "1") return cout<<"1",void();
    if(s == "2") return cout<<"2",void();
    if(!s.size()) return;
    vector<pii> p;
    rep(i,0,999999,1){
        int x = divide(s);
        if(x == -1) break;
        if(!x) continue;
        p.emplace_back(mk(i,1));
    }
    reverse(p.begin(),p.end());
    int len = p.size();
    for(int i = 0;i < len; ++i){
        if(p[i].first == 1) {cout<<2;goto ed;}
        cout<<2;
        if(p[i].first != 1) cout<<'(';
        work(to_string(p[i].first));
        if(p[i].first != 1) cout<<')';
        ed:;
        if(i != len - 1) cout<<'+';
    }
}
inline void solve(){
    cin>>s;
    work(s);
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    solve();
}

子串的子串

赛时写了一个$O(nq\log n)$的暴力，本来是能拿70pts的，但是常数小+信仰就过了。

就是把每次查询的串截出来，然后跑个后缀数组求$height$，用结论 : 一个字符串本质不同的子串的数量为$\frac{n\times (n+1)}{2}-\sum _{i=2}^{n}heigh{t}_i$，就可以了。

求后缀数组用的倍增法，用$DC3$的话就是$O(nq)$的

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = infile("substring.in"),*OutFile = outfile("substring.out");
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 3010;
char *p1,*p2,buf[1<<23];
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<23,stdin),p1==p2)?EOF:*p1++)
#ifdef linux
#define pc putchar_unlocked
#else
#define pc putchar
#endif
namespace IO{
    template<typename T>inline bool read(T &x){x=0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x<<1)+(x<<3)+(s^48);if(!f) x=~x+1;return true;}
    inline bool read(double &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(long double &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(float &x){x=0.0;char s=gc();bool f=true;for(;(s<'0'||'9'<s);s=gc()) {if(s=='-') f=false;if(s==EOF)return false;}for(;'0'<=s&&s<='9';s=gc()) x=(x*10)+(s^48);if(s!='.'){return true;}double res=0.1;s=gc();for(;'0'<=s&&s<='9';res/=10,s=gc()) x+=(s^48)*res;x=f?x:-x;return true;}
    inline bool read(string &str){string ().swap(str);char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false; for(;s!=' '&&s!='\n'&&s!=EOF;s=gc())str.push_back(s);return true;}
    inline bool read_line(string &str){string ().swap(str);char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false;for(;s!='\n'&&s!=EOF;s=gc()){str.push_back(s);}return true;}
    inline bool read_line(char *str){int len=0;char s=gc();for(;s==' '||s=='\n';s=gc());if(s==EOF) return false;for(;s!='\n'&&s!=EOF;s=gc()){str[len]=s;len++;}str[len]='\0';return true;}
    inline bool read(char &s){char x=gc();for(;x==' '||x=='\n';x=gc());if(x==EOF||x==' '||x=='\n')return false;s=x;return true;}
    inline bool read(char *s){int len=0;char x=gc();for(;x==' '||x=='\n';x=gc());if(x==EOF)return false;for(;x!=' '&&x!='\n'&&x!=EOF;x=gc())s[len++]=x;s[len]='\0';return true;}
    template<class T,class... Args> inline bool read(T &x,Args&... args){return (read(x)&&read(args...));}
    template<class T>inline void write(T x){static T st[45];int top=0;if(x<0)x=~x+1,pc('-');do{st[top++]=x%10;}while(x/=10);while(top)pc(st[--top]^48);}
    inline void write(char x){pc(x);}
    inline void write(string s){for(int i=0;s[i];++i) pc(s[i]);}
    inline void write(char *s){int len=strlen(s);for(int i=0;i<len;++i) pc(s[i]);}
    inline void write(const char *s){int len=strlen(s);for(int i=0;i<len;++i) pc(s[i]);}
    template<class T,class... Args> inline void write(T x,Args... args){write(x);write(args...);}
}using namespace IO;
int n,m,sa[N],rk[N],tp[N],ht[N],ct[N];char s[N],s1[N];
inline void Qsort(int m){
    rep(i,1,m,1) ct[i] = 0;
    rep(i,1,n,1) ++ct[rk[i]];
    rep(i,2,m,1) ct[i] += ct[i-1];
    drep(i,n,1,1) sa[ct[rk[tp[i]]]--] = tp[i];
}
inline bool cmp(int x,int y,int w){return (tp[x]==tp[y]&&tp[x+w]==tp[y+w]);}
inline void get_sa(){
    n = strlen(s+1);
    int m = 27;
    rep(i,1,n,1) rk[i] = s[i] - 'a' + 1,tp[i] = i;
    Qsort(m);
    for(int w=1,p=0;w<=n;m=p,p=0,w<<=1){
        rep(i,n-w+1,n,1) tp[++p]=i;
        rep(i,1,n,1) if(sa[i]>w) tp[++p]=sa[i]-w;
        Qsort(m);
        rep(i,1,n,1) swap(tp[i],rk[i]);
        p=rk[sa[1]]=1;
        rep(i,2,n,1) rk[sa[i]]=cmp(sa[i-1],sa[i],w)?p:++p;
        if(p==n) break;
    }
    int k = 0;
    rep(i,1,n,1){
        if(!rk[i]) continue;
        if(k) --k;
        while(s[i+k] == s[sa[rk[i]-1]+k]) k++;
        ht[rk[i]] = k;
    }
}
inline void solve(){
    read(n,m);read(s1+1);
    while(m--){
        int l,r;ll ans;read(l,r);
        rep(i,l,r,1) s[i-l+1] = s1[i];
        n = r-l+1;
        s[n+1] = '\0';
        get_sa();
        ans = 1ll*n*(n+1)/2;
        rep(i,2,n,1) ans -= ht[i];
        rep(i,1,n,1) ht[i] = rk[i] = sa[i] = tp[i] = 0;
        write(ans,'\n');
    }
}
signed main(){
    solve();
}

正解是预处理，二维前缀和。

记$an{s}_{l,r}$表示区间$[l,r]$及其子区间本质不同的字符串的个数。可以先维护差分数组然后前缀和解决。

先枚举子串长度，再枚举左端点，通过预处理的前缀和求出这$[l,r]$一子串的哈希值，那么该串的左右端
点$[l,r]$的答案加一，即$an{s}_{l,r}++$，为了避免重复统计，该串还在上次该串出现位置到该位置
之间该串都有贡献，那么应该将上一次出现的左端点$pre$（字符串哈希维护）与当前右端点的答案减一，
即$an{s}_{pre,r}--$，最后对数组求和得到的其实是会有重复贡献，要通过容斥计算出每个区间的答案$an{s}_{l,r}$
，回答询问

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = infile("substring.in"),*OutFile = outfile("substring.out");
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 3010;
char s[N];
int n,m,ans[N][N];
ull hs[N],pw[N],base = 131;
inline ull get_hash(int l,int r){return hs[r] - hs[l-1]*pw[r-l+1];}
inline void solve(){
    cin>>n>>m>>(s+1);
    pw[0] = 1;rep(i,1,n,1) pw[i] = pw[i - 1] * base;
    rep(i,1,n,1) hs[i] = hs[i-1]*base + s[i];
    rep(len,1,n,1){
        __gnu_pbds::gp_hash_table<ull,int> mp;
        rep(l,1,n-len+1,1){
            int r = l + len - 1;
            ull h = get_hash(l,r);
            ans[mp[h]][r]--;
            mp[h] = l;
        }
    }
    drep(l,n,1,1) rep(r,l,n,1)
        ans[l][r] += ans[l + 1][r] + ans[l][r - 1] - ans[l+1][r-1] + 1;
    while(m--){
        int l,r;cin>>l>>r;cout<<ans[l][r]<<'\n';
    }
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    solve();
}

魔法咒语

考虑一个串是被两个拼起来的或者就一个，一个的特判掉。

考虑对于每一个前缀扩展，假设拓展的第一个字符为$c$

加入$c$已经被扩展过了，那么看看它是否是最后一个节点，如果是，那么说明这个咒语本身就存在，直接加上一个即可。

反之，那么就考虑所有第一个字符为$c$的不同后缀都可以。

前缀的话插入一个正向的Trie就可以，看后缀的话插入一个后缀的Trie即可。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = infile("magic.in"),*OutFile = outfile("magic.out");
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 4e5 + 10;
struct Trie{
    int tree[N][27],cnt[N],tot;
    inline void insert(string s,bool rev){
        int p = 0,len = s.size()-1;
        if(rev) reverse(s.begin(),s.end());
        rep(i,0,len,1){
            int x = s[i] - 'a' + 1;
            if(!tree[p][x]) tree[p][x] = ++tot,cnt[x] += rev;
            p = tree[p][x];
        }
    }
}t1,t2;
ll ans = 0,n;
string s[N];
bitset<27> ok,flag;
inline void solve(){
    cin>>n;
    rep(i,1,n,1){
        cin>>s[i];
        t1.insert(s[i],false);t2.insert(s[i],true);
        int len = s[i].size();
        ok[s[i].back() - 'a' + 1] = true;
        if(len == 1 && !flag[s[i][0] - 'a' + 1]) ans++,flag[s[i][0] - 'a' + 1] = true;
    }
    rep(i,1,t1.tot,1) rep(j,1,26,1)
        if(!t1.tree[i][j]) ans += t2.cnt[j];
        else if(ok[j]) ans++;
    cout<<ans<<'\n';
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    solve();
}

表达式

线段树。不太懂啊。

将给的模数分解成$\prod p_i^{k_i}$，然后以每个$p_i^{k_i}$为模数，求出答案以后再用CRT合并。

至于以$p_i^{k_i}$为模数的处理，用线段树暴力记录就好了。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = infile("expr.in"),*OutFile = outfile("expr.out");
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
#define fi first
#define se second
#define pii pair<char,int>
#define mk make_pair
const int N = 2e5 + 10;
int n,m,mod;
pii a[N];
 
struct Segment_Tree{
    struct segment_tree{
        int l,r,val[30];
        #define l(k) tree[k].l
        #define r(k) tree[k].r
        #define val(k) tree[k].val
    }tree[N<<2];
    int p;
    unordered_map<char,function<int(int,int)> > op = 
    {
        {'+',[&](int a,int b){return (a+b)%p;}},
        {'*',[&](int a,int b){return 1ll*a*b%p;}},
        {'^',[&](int a,int b){int res = 1;for(;b;b >>= 1,a = 1ll*a*a%p) if(b&1) res = 1ll*res*a%p;return res;}}
    };
    inline void pushup(int k){
        rep(i,0,p-1,1) val(k)[i] = val(k<<1|1)[val(k<<1)[i]];
    }
    void build(int k,int l = 1,int r = n){
        l(k) = l,r(k) = r;
        if(l == r){
            rep(i,0,p-1,1) val(k)[i] = op[a[l].fi](i,a[l].se);
            return;
        }
        int mid = (l + r) >> 1;
        build(k<<1,l,mid);build(k<<1|1,mid+1,r);
        pushup(k);
    }
    void change(int k,int pos){
        if(l(k) == r(k)){
            rep(i,0,p-1,1) val(k)[i] = op[a[pos].fi](i,a[pos].se);
            return;
        }
        int mid = (l(k) + r(k)) >> 1;
        if(pos <= mid) change(k<<1,pos);else change(k<<1|1,pos);
        pushup(k);
    }
}T[4];
int exgcd(int a,int b,int &x,int &y){
    if(!b) return x = 1,y = 0,a;
    int d = exgcd(b,a%b,x,y);
    int t = x;x = y;y = t - (a/b)*y;
    return d;
}
namespace TP{
    const int N = 2e5 + 10;
    char op[N];
    int testid,n,m,mod,a[N];
    inline int power(int a,int b,int mod){
        int res = 1;
        for(;b;b>>=1,a = 1ll*a*a%mod) if(b&1) res = 1ll*res*a%mod;
        return res;
    }
    inline void solve(){
        cin>>n>>m>>mod;
        rep(i,1,n,1) cin>>op[i]>>a[i];
        while(m--){
            int p,x;cin>>p;
            if(p == 1){
                cin>>x;
                rep(i,1,n,1){
                    if(op[i] == '+') x = (0ll+x+a[i])%mod;
                    if(op[i] == '*') x = 1ll*x*a[i]%mod;
                    if(op[i] == '^') x = power(x,a[i],mod);
                }
                cout<<x<<'\n';
            }
            else{
                char c;int y;
                cin>>x>>c>>y;op[x] = c,a[x] = y;
            }
        }
    }
}
inline void solve(){
    int id;
    cin>>id;
    if(id <= 3){
        TP::solve();
        return;
    }
    cin>>n>>m>>mod;
    rep(i,1,n,1) cin>>a[i].fi>>a[i].se;
    vector<int> d;
    int res = mod;
    for(int i = 2;i * i <= res; ++i){
        if(res % i == 0){
            int k = 1;
            while(res % i == 0) res /= i,k *= i;
            d.emplace_back(k);
        }
    }
 
    if(res > 1) d.emplace_back(res);
    for(int i = 0;i < d.size(); ++i){
        
        T[i].p = d[i],T[i].build(1);
    }
 
    // exit(0);
 
    auto get_ans = [&](vector<int> &v){
        int res = 0;
        for(int i = 0;i < v.size(); ++i){
            int m = mod/d[i],b,y;
            exgcd(m,d[i],b,y);
            res = (res + 1ll * v[i] * m % mod * b % mod) % mod;
        }
        return (res + mod) % mod;
    };
    
    while(m--){
        int op;cin>>op;
        if(op == 1){
            int x;cin>>x;
            vector<int> v;
            for(int i = 0;i < d.size(); ++i) v.emplace_back(T[i].tree[1].val[x%T[i].p]);
            cout<<get_ans(v)<<'\n';
        }
        else{
            int x,z;cin>>x;char y;cin>>y>>z;
            a[x] = mk(y,z);
            for(int i = 0;i < d.size();T[i++].change(1,x));
        }
    }
 
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    solve();
}