Educational Codeforces Round 75 (Rated for Div. 2) A~E2

Educational Codeforces Round 75 (Rated for Div. 2) A~E2

A. Broken Keyboard

链接

#include<bits/stdc++.h>
using namespace std;
int st[30];
int main()
{
    int T;
    cin >> T;
    while (T--) {
        string s;
        cin >> s;
        for (int i = 0; i < s.length();) {
            if (s[i] == s[i + 1]) {
                i += 2;
                continue;
            }
            st[s[i] - 'a'] = 1;i++;
        }
        for (int i = 0; i < 26; i++) {
            if (st[i]) {
                cout << (char)('a' + i);st[i] = 0;
            }
        }cout << endl;
    }
    return 0;
}

B. Binary Palindromes

链接

#include<bits/stdc++.h>
using namespace std;

void solve()
{
    int n;
    cin >> n;
    int idx = 0, cnt = 0;
    for (int i = 1; i <= n; i++) {
        string s;
        cin >> s;
        int len = s.length();
        if (len % 2) idx++;
        for (int j = 0; j < len; j++) {
            if (s[j] == '1') cnt++;
        }
    }
    if (cnt % 2) {
        if (idx) cout << n << endl;
        else cout << n - 1 << endl;
    }else cout << n << endl;
}

int main()
{
    int T;
    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}

C. Minimize The Integer

链接
题意： 给出一个长度为$3\ast 10^5$的数字，你可以交换相邻的数，但是他们的奇偶性要不同，求能换出来的最小的数字，保留前缀0
思路： 不难看出，同类型不能交换，所以我们顺序存下每个类型的数字，然后输出每个里面更小的那个就行

#include<bits/stdc++.h>
using namespace std;

void solve()
{
    string s;
    cin >> s;
    queue<int> odd, even;
    int len = s.length();
    for (int i = 0; i < len; i++) {
        if ((s[i] - '0') & 1) odd.push(s[i] - '0');
        else even.push(s[i] - '0');
    }
    while (!odd.empty() && !even.empty()) {
        if (odd.front() > even.front()) {cout << even.front();even.pop();}
        else {cout << odd.front();odd.pop();}
    }
    while (!odd.empty()) {cout << odd.front();odd.pop();}
    while (!even.empty()) {cout << even.front();even.pop();}
    cout << endl;
}

int main()
{
    int T;
    T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
}

D. Salary Changing

链接
题意： 给出总共的钱数，和人数，每个人的工资在$l_i和r_i$之间，现在要让他们工资的中位数最大，求出最大的中位数
思路： 二分答案，贪心的求就行

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n, s;
struct Node {int l, r;}a[200010];
bool cmp(Node a, Node b) {
    if (a.l == b.l) return a.r > b.r;
    return a.l > b.l;
}
int all;
bool check(int x)
{
    int num = 0;
    for (int i = 1; i <= n; i++) {
        if (a[i].l > x) num++;
    }
    if (num >= n / 2 + 1) return true;
    int cnt = 0;
    int sum = all;
    if (cnt >= n / 2 + 1) {return true;}
    int idx = cnt + 1;
    while (sum <= s && idx <= n + 1) {
        if (cnt == n / 2 + 1) return true;
        if (idx == n + 1) return false;
        if (a[idx].r < x) {idx++; continue;}
        cnt++; sum += max((x - a[idx].l), 0ll); idx++;
    }return false;
}

void solve()
{
    all = 0;
    cin >> n >> s;
    for (int i = 1; i <= n; i++) {cin >> a[i].l >> a[i].r; all += a[i].l;}
    sort(a + 1, a + 1 + n, cmp);
    int l = 0, r = 1e9+1, ans = 0;
    while (l < r) {
        int mid = l + r >> 1;
        if (check(mid)) {ans = mid, l = mid + 1;}
        else r = mid;
    }
    cout << ans << endl;
    cout << endl;
}

signed main()
{
    int T;
    cin >> T;
    while (T--) solve();
    return 0;
}

E. Voting

链接
题意： 给出每个人的m和p，意思是，你可以花费p购买他，或者等你获得了m个人后自动获取他，问获得全部人最小的花费
思路： 感觉是个有点经典的贪心，首先对于需要人数相同的，一定是先买钱少的，对于需要人数多的，我们可以假设在他之前的所有人已经获得再来获得他，所以对于这种人数，优先买钱少的直到足够多，我们把所有需要购买的放进一个set，每次取小的出来买就行

#include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
    int n;
    cin >> n;
    vector<int> a[n + 1];
    multiset<int> st;
    for (int i = 1; i <= n; i++) {
        int x, y;
        cin >> x >> y;
        a[x].push_back(y);
    }
    int idx = n;
    int buy = 0, ans = 0;
    for (int i = n; i >= 1; i--) {
        idx -= a[i].size();
        int need = i - idx;
        for (int j = 0; j < a[i].size(); j++) {
            st.insert(a[i][j]);
        }
        while(buy < need) {
            ans += (*st.begin());
            st.erase(st.begin());
            buy++;
        }
    }
    cout << ans << endl;
    cout << endl;
}

signed main()
{
    int T;
    cin >> T;
    while (T--) solve();
    return 0;
}