Educational Codeforces Round 70 (Rated for Div. 2)

Educational Codeforces Round 70 (Rated for Div. 2)

A. You Are Given Two Binary Strings...

• 思路：水题

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

int t, len1, len2, pos1, pos2, ans;
string a, b;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        ans = 0;
        cin >> a >> b;
        len1 = a.length(), len2 = b.length();
        for (int i = len2 - 1; i >= 0; i -- ){
            if (b[i] == '1'){
                pos2 = i;
                break;
            }
        }
        for (int i = len1 - (len2 - pos2); i >= 0; i -- ){
            if (a[i] == '1'){
                pos1 = i;
                break;
            }
        }
        cout << (len1 - pos1) - (len2 - pos2) << "
";
    }
    return 0;
}

B. You Are Given a Decimal String...

• 思路：floyd最短路

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 10;
const int INF = 0x3f3f3f3f;

int len, ans;
string s;
bool flag;
int dp[N][N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> s;
    len = s.length();
    for (int x = 0; x < N; x ++ ){
        for (int y = 0; y < N; y ++ ){
            memset(dp, INF, sizeof(dp));
            ans = 0;
            flag = true;
            for (int i = 0; i < N; i ++ ){
                dp[i][(i + x) % 10] = 1;
                dp[i][(i + y) % 10] = 1;
            }
            for (int k = 0; k < N; k ++ )
                for (int i = 0; i < N; i ++ )
                    for (int j = 0; j < N; j ++ )
                        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
            /* for (int i = 0; i < N; i ++ ){
                for (int j = 0; j < N; j ++ )
                    cout << dp[i][j] << " ";
                cout << "
";
            } */
            for (int i = 1; i < len; i ++ ){
                int l = s[i - 1] - '0', r = s[i] - '0';
                if (dp[l][r] == INF){
                    flag = false;
                    break;
                }
                ans += dp[l][r] - 1;
            }
            if (flag)
                cout << ans << " ";
            else
                cout << -1 << " ";
        }
        cout << "
";
    }
    return 0;
}

D. Print a 1337-string...（赛后补）

• 思路：构造出类似于133/77...77/33...33/7这样的串

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

int t, n, cnt;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        cnt = 1;
        cin >> n;
        cout << "133";
        while (cnt * (cnt - 1) / 2 <= n)
            cnt ++ ;
        cnt -- ;
        for (int i = 1; i <= n - cnt * (cnt - 1) / 2; i ++ )
            cout << "7";
        for (int i = 1; i <= cnt - 2; i ++ )
            cout << "3";
        cout << "7
";
    }
    return 0;
}

E. You Are Given Some Strings...

• 思路：正着反着分别跑一次AC自动机

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 2e5 + 10;

struct Trie{
    int next[N][26], fail[N], end[N], res[N];
    int tot;
    int newnode(){
        int now = ++ tot;
        end[now] = 0;
        return now;
    }
    void init(){
        tot = 0;
        newnode();
    }
    void insert(string buf){
        int len = buf.length();
        int now = 1;
        for (int i = 0; i < len; i ++ ){
            if (!next[now][buf[i] - 'a'])
                next[now][buf[i] - 'a'] = newnode();
            now = next[now][buf[i] - 'a'];
        }
        end[now] ++ ;
    }
    void build(){
        queue<int> q;
        for (int i = 0; i < 26; i ++ ){
            if (!next[1][i])
                next[1][i] = 1;
            else{
                fail[next[1][i]] = 1;
                q.push(next[1][i]);
            }
        }
        while(!q.empty()){
            int now = q.front();
            q.pop();
            end[now] += end[fail[now]];
            for (int i = 0; i < 26; i ++ ){
                if (!next[now][i])
                    next[now][i] = next[fail[now]][i];
                else{
                    fail[next[now][i]] = next[fail[now]][i];
                    q.push(next[now][i]);
                }
            }
        }
    }
}ac1, ac2;

int n;
ll ans;
string s, t;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> s >> n;
    int len = s.length();
    ac1.init(), ac2.init();
    for (int i = 1; i <= n; i ++ ){
        cin >> t;
        // cout << t << " ";
        ac1.insert(t);
        reverse(t.begin(), t.end());
        // cout << t << " ";
        ac2.insert(t);
    }
    ac1.build(), ac2.build();
    for (int i = 0, now = 1; i < len; i ++ ){
        ac1.res[i] = ac1.end[ac1.next[now][s[i] - 'a']];
        now = ac1.next[now][s[i] - 'a'];
    }
    for (int i = len - 1, now = 1; i >= 0; i -- ){
        ac2.res[i] = ac2.end[ac2.next[now][s[i] - 'a']];
        now = ac2.next[now][s[i] - 'a'];
    }
    for (int i = 0; i < len - 1; i ++ )
        ans += 1ll * ac1.res[i] * ac2.res[i + 1];
    cout << ans << "
";
    return 0;
}