Educational Codeforces Round 39 

D. Timetable

令\(dp[i][j]\)表示前\(i\)天逃课了\(j\)节课的情况下，在学校的最少时间

转移就是枚举第\(i\)天逃了\(x\)节课，然后取当天逃\(x\)节课情况下在学校的最小值即可

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 2e5+7;

int n, m, k;

char s[MAXN];

void solve(){

    sci(n); sci(m); sci(k);

    vi f(k+1,n*m);

    f[k] = 0;

    for(int d = 1; d <= n; d++){

        vi next_f(k+1,n*m);

        scs(s+1);

        vi A; for(int i = 1; i <= m; i++) if(s[i]=='1') A << i;

        vi cost(A.size()+1,n*m);

        for(int i = 0; i <= A.size(); i++){

            if(i==A.size()) cost[i] = 0;

            else for(int j = 0; j <= i; j++) cmin(cost[i],A[A.size() - 1 - i + j] - A[j] + 1);

        }

        for(int pre = 0; pre <= k; pre++) for(int cur = 0; cur <= min(pre,(int)A.size()); cur++) cmin(next_f[pre-cur],f[pre]+cost[cur]);

        f.swap(next_f);

    }

    cout << *min_element(all(f)) << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    solve();

    return 0;

}

E.  Largest Beautiful Number

考虑从后往前枚举每个位置，判断在这个位置之前全部相同，这个位置比原来小的情况下是否存在合法解即可

特判长度为奇数的情况和小于等于\(1xxxxxx1\)的情况

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 2e5+7;

int n;

string s;

void solve(){

    cin >> s;

    function<bool(void)>check = [&](){

        string str(s.size(),'0');

        str.front() = str.back() = '1';

        return str >= s;

    };

    if(s.size()&1 or check()){

        for(int i = 1; i <= (s.size()&1 ? s.size() - 1 : s.size() - 2); i++) cout << 9;

        cout << endl; return;

    }

    vi cnt(10,0);

    for(int i = 0; i < s.size(); i++) cnt[s[i]-'0'] ^= 1;

    for(int i = s.size() - 1; ~i; i--){

        cnt[s[i]-'0'] ^= 1;

        for(char c = s[i] - 1; c >= '0'; c--){

            cnt[c-'0'] ^= 1;

            int tot = accumulate(all(cnt),0);

            if(tot>s.size()-i-1){

                cnt[c-'0'] ^= 1;

                continue;

            }

            string str(s);

            s[i] = c;

            int cur = i;

            for(int j = 0; j < s.size()-i-1-tot; j++) s[++cur] = '9';

            for(int j = 9; ~j; j--) if(cnt[j]) s[++cur] = j + '0';

            cout << s << endl;

            return;

        }

    }

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    int tt; for(cin >> tt; tt--; solve());

    return 0;

}

F. Fibonacci String Subsequences

\(dp[i][l][r]\)表示\(f(i)\)中子串\(s[l:r]\)以子序列的方式出现的次数，考虑转移

对于一般情况，可以考虑把\(s[l:r]\)分成\(s[l:k]\)和\(s[k+1:r]\)，然后分别在\(f(i-1)\)和\(f(i-2)\)中找

其次考虑\(s[l:r]\)全在\(f(i-1)\)或\(f(i-2)\)中，如果\(r=n-1\)，那么在\(f(i-1)中找到\)\(s[l:r]\)出现次数之后由于后面可以随便选，乘上\(2^{fib(i-2)}\)即可，否则只算左边一部分，对于\(l=0\)的情况类似处理即可

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 2e2+7;

const int MOD = 1e9+7;

LL ksm(LL a, LL b){

    LL ret = 1;

    while(b){

        if(b&1) ret = ret * a % MOD;

        b >>= 1;

        a = a * a % MOD;

    }

    return ret;

}

int n, x, fib[MAXN], f[MAXN][MAXN][MAXN];

string s;

int dp(int x, int l, int r){

    int &ret = f[x][l][r];

    if(~ret) return ret;

    if(x==0) return (ret = ((l==r) and s[l]=='0'));

    if(x==1) return (ret = ((l==r) and s[l]=='1'));

    ret = 0;

    if(r==n-1) ret = (ret + 1ll * dp(x-1,l,r) * ksm(2,fib[x-2])) % MOD;

    else ret = (ret + 1ll * dp(x-1,l,r)) % MOD;

    if(l==0) ret = (ret + 1ll * dp(x-2,l,r) * ksm(2,fib[x-1])) % MOD;

    else ret = (ret + 1ll * dp(x-2,l,r)) % MOD;

    for(int i = l; i < r; i++) ret = (ret + 1ll * dp(x-1,l,i) * dp(x-2,i+1,r)) % MOD;

    return ret;

}

void solve(){

    cin >> n >> x >> s;

    fib[0] = fib[1] = 1;

    for(int i = 2; i < MAXN; i++) fib[i] = (fib[i-1] + fib[i-2]) % (MOD - 1);

    memset(f,255,sizeof(f));

    cout << dp(x,0,n-1) << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    solve();

    return 0;

}

G. Almost Increasing Array

假设没有删一个数的条件的话，就是把原序列\(A\)中的所有\(A[i]\)变成\(A[i]-i\)，然后算最长非降子序列

现在考虑枚举删掉的数为\(A[k]\)，那么对于\(k\)之前的数\(A[i]\)来说，需要减去\(i\)，对于\(k\)之后的数\(A[j]\)来说，需要减去\(j-1\)

那么之需要知道以\(k-1\)为结尾的最长非降子序列的长度和大于\(k\)的部分中起始值大于等于\(A[k-1]\)的最长非降子序列即可

那么可以用线段树来维护

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 4e5+7;

int n, A[MAXN];

struct SegmentTree{

    int l[MAXN<<2], r[MAXN<<2], maxx[MAXN<<2];

    #define ls(rt) rt << 1

    #define rs(rt) rt << 1 | 1

    #define pushup(rt) maxx[rt] = max(maxx[ls(rt)],maxx[rs(rt)])

    void build(int L, int R, int rt = 1){

        l[rt] = L; r[rt] = R;

        maxx[rt] = 0;

        if(L+1==R) return;

        int mid = (L + R) >> 1;

        build(L,mid,ls(rt)); build(mid,R,rs(rt));

    }

    void modify(int pos, int x, int rt = 1){

        if(l[rt] + 1 == r[rt]){

            maxx[rt] = x;

            return;

        }

        int mid = (l[rt] + r[rt]) >> 1;

        if(pos<mid) modify(pos,x,ls(rt));

        else modify(pos,x,rs(rt));

        pushup(rt);

    }

    int qmax(int L, int R, int rt = 1){

        if(L>=r[rt] or l[rt]>=R) return 0;

        if(L<=l[rt] and r[rt]<=R) return maxx[rt];

        return max(qmax(L,R,ls(rt)),qmax(L,R,rs(rt)));

    }

}ST;

int f[MAXN];

void solve(){

    sci(n);

    vi vec;

    for(int i = 1; i <= n; i++) sci(A[i]), A[i] -= i, vec << A[i] << A[i] + 1;

    sort(all(vec)); vec.erase(unique(all(vec)),vec.end());

    auto ID = [&](int x){ return lower_bound(all(vec),x) - vec.begin() + 1; };

    ST.build(0,vec.size()+1);

    for(int i = 1; i <= n; i++){

        int x = ID(A[i]);

        f[i] = ST.qmax(0,x+1) + 1;

        ST.modify(x,f[i]);

    }

    int ret = f[n-1];

    ST.build(0,vec.size()+1);

    for(int i = n; i >= 2; i--){

        ret = max(ret,f[i-1]+ST.qmax(ID(A[i-1]),vec.size()+1));

        int x = ID(A[i]+1);

        ST.modify(x,ST.qmax(x,vec.size()+1)+1);

    }

    cmax(ret,ST.qmax(0,vec.size()+1));

    cout << n - 1 - ret << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    solve();

    return 0;

}

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