Codeforces Round 878 (Div. 3)

~~A：ABC~~

A. Cipher Shifer

题意：在自身后面添加一个字母，但是不能添加自身

思路：找到第二个与自身相符的就再找

#include <bits/stdc++.h>

using namespace std;
const int MAX = 110;
char a[MAX];

void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    int num = 0;
    for (int i = 1; i < n; i++) {
        if (a[num] == a[i]) {
            cout << a[num];
            num = ++i;
        }
    }
    cout << "\n";
}

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

B. Binary Cafe

题意：求一个数n用<2^k的二进制数的表示方法

思路：每个数都有一个独立的二进制来表示，每个二进制表示都可记作一种方案（空集是一种单独的方案）int>1e9所以k>32直接输出n+1就行。其他的再判断就行

#include <bits/stdc++.h>

using namespace std;
#define int long long

int qmi(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) {
            res = res * a;
        }
        a *= a;
        b >>= 1;
    }
    return res;
}

void solve() {
    int n, k;
    cin >> n >> k;
    if (k >= 32) {
        cout << n + 1 << "\n";
    } else {
        int res = qmi(2, k) - 1;
        if (res > n) cout << n + 1 << "\n";
        else {
            cout << res + 1 << "\n";
        }
    }
}

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

C. Ski Resort

题意：一共n天，要连续去k天，这k天中a[i]<=q

思路：用一个记录有多少天连续满足这个条件，然后如果cnt>=k那么急可以在这几天去

#include <bits/stdc++.h>

using namespace std;
#define int long long
const int MAX = 2e5 + 10;
bool t[MAX];
int dp[MAX];

void solve() {
    int n, k, q;
    cin >> n >> k >> q;
    for (int i = 1; i <= n; i++) {
        int x;
        cin >> x;
        if (x <= q) {
            t[i] = true;
        } else t[i] = false;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        if (t[i]) cnt++;
        else cnt = 0;
        dp[i] = dp[i - 1];
        if (cnt >= k) {
            dp[i] += cnt - k + 1;
        }
    }
    cout << dp[n] << "\n";

}

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

D. Wooden Toy Festival

题意：有x,y,z三个数，要让abs(a[i]-x)+abs(a[i]-y)+abs(a[i]-z)最小

思路：二分答案，是否有满足大于当前二分答案的两倍的个数大于3（大致分成三份），如果有就在大于当前二分的值的区间，否则就左边找

#include <bits/stdc++.h>

using namespace std;
#define int long long
const int MAX = 2e5 + 10;
int a[MAX];
int n;

bool check(int k) {
    int num = 0;
    int x = a[1];
    for (int i = 1; i <= n; i++)
        if (abs(a[i] - x) > k * 2) {
            x = a[i];
            num++;
        }
    if (num >= 3) return false;
    return true;
}

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    if (n <= 3) {
        cout << "0" << endl;
        return;
    }
    sort(a + 1, a + n + 1);
    int l = 0, r = a[n];
    int mid;
    while (l <= r) {
        mid = (l + r) >> 1;
        if (check(mid)) r = mid - 1;
        else l = mid + 1;
    }
    cout << l << "\n";
}

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

E. Character Blocking

总结：大致思路有，但是没想到用map来存麻痹时间，题刷少了

题意：有三种操作:1.在 t 秒内屏蔽两个字符串中位于 pos位置的字符

2.交换两个未屏蔽的字符（注：不一定是不同字符串的）

3.确定两个字符在查询是是否相等

思路：找一个cnt确定有几个数是坏(没被屏蔽并且不相等)，用map来储存屏蔽时间

#include <bits/stdc++.h>

using namespace std;

void solve() {
    string a, b;
    cin >> a >> b;
    a = " " + a;
    b = " " + b;
    int t, m;
    cin >> t >> m;
    int cnt = 0;
    map<int, vector<int>> mp;
    for (int i = 1; i < a.size(); i++) {
        cnt += a[i] != b[i];
    }
    for (int i = 1; i <= m; i++) {//屏蔽时间问题
        for (auto &x: mp[i]) {
            if (a[x] != b[x]) cnt++;
        }
        int op;
        cin >> op;
        if (op == 1) {
            int x;
            cin >> x;
            cnt -= (a[x] != b[x]);
            mp[i + t].push_back(x);//i+t是解除屏蔽的时间
        } else if (op == 2) {
            int x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;
            if (x1 == 1 && x2 == 1) {
                if (a[y1] != b[y1]) cnt--;
                if (a[y2] != b[y2]) cnt--;
                swap(a[y1], a[y2]);
                if (a[y1] != b[y1]) cnt++;
                if (a[y2] != b[y2]) cnt++;
            } else if (x1 == 1 && x2 == 2) {
                if (a[y1] != b[y1]) cnt--;
                if (a[y2] != b[y2]) cnt--;
                swap(a[y1], b[y2]);
                if (a[y1] != b[y1]) cnt++;
                if (a[y2] != b[y2]) cnt++;
            } else if (x1 == 2 && x2 == 1) {
                if (b[y1] != a[y1]) cnt--;
                if (b[y2] != a[y2]) cnt--;
                swap(b[y1], a[y2]);
                if (b[y1] != a[y1]) cnt++;
                if (b[y2] != a[y2]) cnt++;
            } else if (x1 == 2 && x2 == 2) {
                if (b[y1] != a[y1]) cnt--;
                if (b[y2] != a[y2]) cnt--;
                swap(b[y1], b[y2]);
                if (b[y1] != a[y1]) cnt++;
                if (b[y2] != a[y2]) cnt++;
            }
        } else {
            if (cnt) cout << "NO\n";
            else cout << "YES\n";
        }
    }
}

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