Educational Codeforces Round 30 

A. Chores

把最大的换掉

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("%I64d",&x)

#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;

void solve(){

    int n, m, k;

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

    vi A(n); for(int &x : A) sci(x);

    cmin(m,n);

    for(int i = n - 1, j = 0; j < m; j++, i--) cmin(A[i],k);

    cout << accumulate(all(A),0) << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

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

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

    #endif

    solve();

    return 0;

}

B.  Balanced Substring

令\(s\)为前缀和

就是要找\(s_r-s_{l-1}=\frac {r-l+1}{2}\)

那就是\(2s_r-r = 2s_{l-1}-(l-1)\)

维护每个值最早出现的位置

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("%I64d",&x)

#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;

char s[MAXN];

void solve(){

    int n; cin >> n;

    cin >> s + 1;

    map<int,int> msk;

    int pre = 0;

    msk[0] = 0;

    int ret = 0;

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

        pre += s[i] - '0';

        if(msk.count(2*pre-i)) cmax(ret,i-msk[2*pre-i]);

        else msk[2*pre-i] = i;

    }

    cout << ret << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

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

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

    #endif

    solve();

    return 0;

}

C. Strange Game On Matrix

每一列单独考虑，枚举\(q\)的起点

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("%I64d",&x)

#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 A[111][111];

void solve(){

    int n, m, k;

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

    for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) sci(A[i][j]);

    int ret = 0, exc = 0;

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

        vi pos;

        for(int j = 1; j <= n; j++) if(A[j][i]==1) pos << j;

        if(pos.empty()) continue;

        int tmpret = 0, tmpexc = 0;

        for(int j = 0; j < (int)pos.size(); j++){

            int sum = 0;

            for(int kk = j; kk < pos.size(); kk++){

                if(pos[kk] - pos[j] >= k) break;

                sum++;

            }

            if(sum > tmpret) tmpret = sum, tmpexc = j;

        }

        ret += tmpret; exc += tmpexc;

    }

    cout << ret << ' ' << exc << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

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

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

    #endif

    solve();

    return 0;

}

D. Merge Sort

递归下去，如果还需要调用的话，就把当前数字区间左右互换然后分别调用，否则直接不换分别调用

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("%I64d",&x)

#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, k, A[MAXN];

void merge(int L, int R, int numl, int numr, int &T){

    if(L+1==R){

        A[L] = numl;

        return;

    }

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

    if(T>0) T-=2, merge(L,mid,numr-mid+L,numr,T), merge(mid,R,numl,numr-mid+L,T);

    else merge(L,mid,numl,numl+mid-L,T), merge(mid,R,numl+mid-L,numr,T);

}

void solve(){

    sci(n); sci(k); k--;

    merge(0,n,1,n+1,k);

    if(k!=0) cout << -1 << endl;

    else for(int i = 0; i < n; i++) cout << A[i] << ' ';

}

int main(){

    #ifndef ONLINE_JUDGE

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

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

    #endif

    solve();

    return 0;

}

E. Awards For Contestants

先所有数从大到小排序

枚举前两个的位置然后可以确定下一个位置的可行范围，然后\(ST\)表找最大的位置即可

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("%I64d",&x)

#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 = 3333;

int n, d1, d2, d3, x, y, z, ret[MAXN];

pii A[MAXN];

pii st[MAXN][20];

void solve(){

    sci(n); for(int i = 1; i <= n; i++) sci(A[i].first), A[i].second = i;

    sort(A+1,A+1+n,greater<pii>());

    for(int i = 1; i <= n; i++) st[i][0] = make_pair(A[i].first-A[i+1].first,i);

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

        if(st[i][j-1].first > st[i+(1<<(j-1))][j-1].first) st[i][j] = st[i][j-1];

        else st[i][j] = st[i+(1<<(j-1))][j-1];

    }

    auto query = [&](int l, int r){

        int d = (int)log2(r - l + 1);

        if(st[l][d].first>st[r-(1<<d)+1][d].first) return st[l][d];

        else return st[r-(1<<d)+1][d];

    };

    d1 = d2 = d3 = -1;

    for(int i = 1; i <= n - 2; i++) for(int j = i + 1; j <= n - 1; j++){

        int a = i, b = j - i;

        if(min(a,b) * 2 < max(a,b)) continue;

        int l = 1, r = n - j;

        cmax(l,(max(a,b)+1)/2);

        cmin(r,min(a,b)*2);

        if(l>r) continue;

        l += j; r += j;

        auto p = query(l,r);

        int t1 = A[i].first - A[i+1].first, t2 = A[j].first - A[j+1].first, t3 = p.first;

        if(t1>d1 or (t1==d1 and t2>d2) or (t1==d1 and t2==d2 and t3>d3)){

            d1 = t1; d2 = t2; d3 = t3;

            x = i; y = j; z = p.second;

        }

    }

    for(int i = 1; i <= x; i++) ret[A[i].second] = 1;

    for(int i = x + 1; i <= y; i++) ret[A[i].second] = 2;

    for(int i = y + 1; i <= z; i++) ret[A[i].second] = 3;

    for(int i = z + 1; i <= n; i++) ret[A[i].second] = -1;

    for(int i = 1; i <= n; i++) cout << ret[i] << ' ';

    cout << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

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

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

    #endif

    solve();

    return 0;

}

F. Forbidden Indices

后缀自动机，插入一个字符的时候如果这个位置末尾被禁止的话，当前点的\(right\)集合大小设为\(0\)，否则设为\(1\)，然后就是计算每个状态节点的\(len[i]\cdot right[i]\)最大值了

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("%I64d",&x)

#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 = 1e6+7;

char s[MAXN], t[MAXN];

struct SAM{

    int len[MAXN],link[MAXN],ch[MAXN][26],cnt[MAXN],tot,last;

    int buc[MAXN], sa[MAXN];

    SAM(){ link[0] = -1; }

    void extend(int c, int ct){

        int np = ++tot, p = last;

        len[np] = len[last] + 1; cnt[np] = ct;

        while(p!=-1 and !ch[p][c]){

            ch[p][c] = np;

            p = link[p];

        }

        if(p==-1) link[np] = 0;

        else{

            int q = ch[p][c];

            if(len[p]+1==len[q]) link[np] = q;

            else{

                int clone = ++tot;

                len[clone] = len[p] + 1;

                link[clone] = link[q];

                memcpy(ch[clone],ch[q],sizeof(ch[q]));

                link[np] = link[q] = clone;

                while(p!=-1 and ch[p][c]==q){

                    ch[p][c] = clone;

                    p = link[p];

                }

            }

        }

        last = np;

    }

    void rua(){

        for(int i = 1; i <= tot; i++) buc[i] = 0;

        for(int i = 1; i <= tot; i++) buc[len[i]]++;

        for(int i = 1; i <= tot; i++) buc[i] += buc[i-1];

        for(int i = tot; i >= 1; i--) sa[buc[len[i]]--] = i;

        LL ret = 0;

        for(int i = tot; i >= 1; i--){

            int u = sa[i];

            cnt[link[u]] += cnt[u];

        }

        for(int i = 1; i <= tot; i++) cmax(ret, 1ll * cnt[i] * len[i]);

        cout << ret << endl;

    }

}sam;

void solve(){

    int len;

    cin >> len >> s >> t;

    for(int i = 0; i < len; i++) sam.extend(s[i]-'a',(t[i]-'0')^1);

    sam.rua();

}

int main(){

    #ifndef ONLINE_JUDGE

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

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

    #endif

    solve();

    return 0;

}

Educational Codeforces Round 30的更多相关文章

  1. Educational Codeforces Round 30 D. Merge Sort

    题意:给你n和k,n代表有多少个数,k代表几次操作,求一个1到n的序列,要k次mergesort操作才能还原 Examples Input 3 3 Output 2 1 3 Input 4 1 Out ...

  2. Educational Codeforces Round 30 B【前缀和+思维/经典原题】

    B. Balanced Substring time limit per test 1 second memory limit per test 256 megabytes input standar ...

  3. Educational Codeforces Round 30 A[水题/数组排序]

    A. Chores time limit per test 2 seconds memory limit per test 256 megabytes input standard input out ...

  4. Educational Codeforces Round 53 E. Segment Sum(数位DP)

    Educational Codeforces Round 53 E. Segment Sum 题意: 问[L,R]区间内有多少个数满足:其由不超过k种数字构成. 思路: 数位DP裸题,也比较好想.由于 ...

  5. Educational Codeforces Round 76 (Rated for Div. 2) E. The Contest

    Educational Codeforces Round 76 (Rated for Div. 2) E. The Contest(dp+线段树) 题目链接 题意: 给定3个人互不相同的多个数字,可以 ...

  6. Educational Codeforces Round 85 (Rated for Div. 2)

    \(Educational\ Codeforces\ Round\ 85\ (Rated\ for\ Div.2)\) \(A. Level Statistics\) 每天都可能会有人玩游戏,同时一部 ...

  7. Educational Codeforces Round 117 (Rated for Div. 2)

    Educational Codeforces Round 117 (Rated for Div. 2) A. Distance https://codeforces.com/contest/1612/ ...

  8. [Educational Codeforces Round 16]E. Generate a String

    [Educational Codeforces Round 16]E. Generate a String 试题描述 zscoder wants to generate an input file f ...

  9. [Educational Codeforces Round 16]D. Two Arithmetic Progressions

    [Educational Codeforces Round 16]D. Two Arithmetic Progressions 试题描述 You are given two arithmetic pr ...

随机推荐

  1. Gradle最佳实践

    一.Gradle相比Maven的优势 配置简洁 Maven是用pom.xml管理,引入一个jar包至少5行代码,Gradle只需要一行. 构建速度快 Gradle支持daemon方式运行,启动速度快, ...

  2. Java 多线程读取文件并统计词频 实例 出神入化的《ThreadPoolExecutor》

    重在展示多线程ThreadPoolExecutor的使用,和线程同步器CountDownLatch,以及相关CAS的原子操作和线程安全的Map/队列. ThreadPool主线程 1 import j ...

  3. GitLab-CI/CD入门实操

    以Spring boot项目为例.传统方式是本地生成jar包,FTP上传服务器,重启服务:如果是内网测试服,也可以在服务器上安装git,在服务器上编译打包.但这都需要人为干预,于是CI/CD就出现了. ...

  4. mysql+MHA高可用

    MHA(Master High Availability)目前在MySQL高可用方面是一个相对成熟的解决方案,它由日本DeNA公司youshimaton(现就职于Facebook公司)开发,是一套优秀 ...

  5. OpenCV 和 Dlib 人脸识别基础

    00 环境配置 Anaconda 安装 1 下载 https://repo.anaconda.com/archive/ 考虑到兼容性问题,推荐下载Anaconda3-5.2.0版本. 2 安装 3 测 ...

  6. Wi-Fi IoT套件连PCF8563实现电子钟功能

    首先跟同样新入手单片机开发的小伙伴分享一点I2C通信的知识.我估计大部分入手开发板的小伙伴都有一定程序开发的能力,但是底层开发可能是新接触,我看有的小伙伴配置开发环境都有障碍,其实并不是多复杂,只是首 ...

  7. Ajax编程基础

    目录 Ajax编程基础 传统网站中存在的问题 Ajax概述 Ajax的应用场景 Ajax的运行环境 Ajax运行原理及实现 Ajax运行原理 Ajax的实现步骤 1.创建Ajax对象 2.告诉Ajax ...

  8. Oracle 0至6级锁的通俗解释及实验案例_ITPUB博客 http://blog.itpub.net/30126024/viewspace-2156232/

    Oracle 0至6级锁的通俗解释及实验案例_ITPUB博客 http://blog.itpub.net/30126024/viewspace-2156232/

  9. Connection reset by peer的常见原因及解决办法 RST 大文件上传

    Connection reset by peer的常见原因及解决办法 Connection reset by peer的常见原因 - 简书 https://www.jianshu.com/p/263e ...

  10. Ubuntu16 安装 OpenSSH-Server

    Ubuntu16.04 桌面版默认是没有安装 SSH 服务的,需要手动安装服务: 更新源:sudo apt-get update 安装服务:sudo apt-get install -y openss ...