Codeforces Round #595 (Div. 3)

Codeforces Round #595 (Div. 3)

A. Yet Another Dividing into Teams

• 思路：排序后找是否有两个相邻的数差值为1即可

• 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 = 110;

int q, n;
int a[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 >> q;
    while (q -- ){
        cin >> n;
        for (int i = 1; i <= n; i ++ )
            cin >> a[i];
        sort(a + 1, a + n + 1);
        bool flag = false;
        for (int i = 1; i < n; i ++ ){
            if (a[i] == a[i + 1] - 1){
                flag = true;
                break;
            }
        }
        if (flag)
            cout << 2 << "\n";
        else
            cout << 1 << "\n";
    }
    return 0;
}

B2. Books Exchange (hard version)

• 思路：dfs找环

• 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;
}

ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

const int N = 2e5 + 10;

int q, n;
int p[N], ans[N];

int calc(int st, int now, int to, int step){
    if (st == to)
        return ans[now] = step;
    return ans[now] = calc(st, to, p[to], step + 1);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> q;
    while (q -- ){
        memset(ans, 0, sizeof(ans));
        cin >> n;
        for (int i = 1; i <= n; i ++ )
            cin >> p[i];
        for (int i = 1; i <= n; i ++ )
            if (!ans[i])
                calc(i, i, p[i], 1);
        for (int i = 1; i <= n; i ++ ){
            if (i != n)
                cout << ans[i] << " ";
            else
                cout << ans[i] << "\n";
        }
    }
    return 0;
}

C2. Good Numbers (hard version)

• 思路：转换为三进制 每一位\(\geq2\)的当前位清零 下一位\(+1\) 并标记 然后把标记位之前的全部清零 再转换成十进制就行了 没写标记位直接将最高位之前清零导致wa6（逃

• 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 = 100;

int q;
ll n, ans;
int t[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 >> q;
    while (q -- ){
        memset(t, 0, sizeof(t));
        ans = 0;
        cin >> n;
        int cnt = 0;
        ll tmp = n;
        while (tmp){
            tmp /= 3;
            t[cnt ++ ] = n - 3 * tmp;
            n = tmp;
        }
        int flag = -1;
        for (int i = 0; i < cnt; i ++ ){
            if (t[i] >= 2){
                flag = i;
                t[i] = 0;
                t[i + 1] ++ ;
            }
        }
        if (flag != -1){
            for (int i = 0; i < flag; i ++ )
                t[i] = 0;
        }
        ll fac = 1;
        for (int i = 0; i <= cnt; i ++ ){
            ans += t[i] * fac;
            fac *= 3;
        }
        cout << ans << "\n";
    }
    return 0;
}

D2. Too Many Segments (hard version)

• 思路：看了其他dalao博客的思路 将所有线段用set维护 然后从小到大枚举端点 将左端点覆盖到这个点的线段加入set中 然后线段数量大于k时移除右端点最大的线段 最后将右端点等于这个点的线段全部移除 （太妙了

• 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;
}

ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

typedef pair<ll, ll> pii;

const int N = 2e5 + 10;

int n, k, l, r;
vector<pii> vec[N];
vector<int> ans;
set<pii> st;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; i ++ ){
        cin >> l >> r;
        vec[l].push_back({r, i});
    }
    for (int i = 0; i < N; i ++ ){
        for (int j = 0; j < vec[i].size(); j ++ )
            st.insert(vec[i][j]);
        while (st.size() > k){
            ans.push_back(st.rbegin() -> second);
            st.erase( -- st.end());
        }
        while (!st.empty() && st.begin() -> first == i){
            st.erase(st.begin());
        }
    }
    cout << ans.size() << "\n";
    for (int i = 0; i < ans.size(); i ++ ){
        if (i != ans.size() - 1)
            cout << ans[i] << " ";
        else
            cout << ans[i] << "\n";
    }
    return 0;
}

/*
7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9

3
7 4 6
*/

/*
5 1
29 30
30 30
29 29
28 30
30 30


3
4 1 5
*/

/*
6 1
2 3
3 3
2 3
2 2
2 3
2 3

4
6 5 3 1
*/

E. By Elevator or Stairs?

• 思路：比较简单的dp 详细见代码

• 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;
}

ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

const int N = 2e5 + 10;
const int M = 3;

int n, c;
int a[N], b[N];
int dp[N][M];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> c;
    memset(dp, 0x3f, sizeof(dp));
    dp[1][1] = 0, dp[1][2] = c;
    for (int i = 1; i < n; i ++ )
        cin >> a[i];
    for (int i = 1; i < n; i ++ )
        cin >> b[i];
    for (int i = 1; i < n; i ++ ){
        dp[i + 1][1] = min(dp[i + 1][1], dp[i][1] + a[i]);
        dp[i + 1][1] = min(dp[i + 1][1], dp[i][2] + a[i]);
        dp[i + 1][2] = min(dp[i + 1][2], dp[i][1] + c + b[i]);
        dp[i + 1][2] = min(dp[i + 1][2], dp[i][2] + b[i]);
    }
    for (int i = 1; i <= n; i ++ ){
        if (i != n)
            cout << min(dp[i][1], dp[i][2]) << " ";
        else
            cout << min(dp[i][1], dp[i][2]) << "\n";
    }
    return 0;
}
/*
0 2
7 8
13 17
31 18
24 28
40 35
53 36
37 46
54 40
57 45
*/
/*
0 1
3 2
4 4
7 7
8 11
11 13
14 13
14 15
18 16
17 19
*/