Codeforces Round #826 (Div. 3) A-E

Codeforces Round #826 (Div. 3)

https://codeforces.com/contest/1741

今天vp了一下最近一场div3，感觉难度在上升啊（悲
vp出了abd，c是我sb了有个地方初始化错了。
然后预期的话应该是要能出E的（“简单dp”）
FG还没补，感觉小难
F是一个动态开点线段树，周末学学再补。
G是一个复杂dp呜

A. Compare T-Shirt Sizes

嗯模拟

#include <bits/stdc++.h>

using namespace std;
map <char, int> mp;

void solve () {
    string a, b;
    cin >> a >> b;
    if (a == b) {
        puts ("=");
        return;
    }
    int n = a.size (), m = b.size ();
    for (int i = n - 1, j = m - 1; i >= 0 && j >= 0; i--, j--) {
        if (a[i] == b[j])   continue;
        else {
            if (mp[a[i]] > mp[b[j]])    puts (">");
            else    puts ("<");
            return ;
        }
    }
    if (a[n-1] == 'S') {
        if (n < m)  puts (">");
        else    puts ("<");
    }
    else {
        if (n > m)  puts (">");
        else    puts ("<");
    }

}

int main () {
    mp['S'] = 0, mp['M'] = 1, mp['L'] = 2;
    int t;
    cin >> t;
    while (t --)    solve ();
}

//S, M, L

B. Funny Permutation

简单构造

#include <bits/stdc++.h>

using namespace std;

void solve () {
    int n;
    cin >> n;
    if (n == 3) puts ("-1");
    else {
        if (n & 1) {
            for (int i = (n + 1) / 2 + 1; i <= n; i++)  cout << i << ' ';
            for (int i = 1; i <= (n + 1) / 2; i++)  cout << i << ' ';
            cout << endl;
        }
        else {
            for (int i = n; i >= 1; i--)    cout << i << ' ';
            cout << endl;
        }
    }
}

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

//S, M, L

C. Minimize the Thickness

枚举第一段的长度然后贪心构造

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 2005;
int a[N], n, ans, sum;

void solve () {
    cin >> n;
    ans = n, sum = 0;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    for (int len = 1; len < n; len++) {
        //后面贪心构造
        sum += a[len];
        //cout << "sum=" << sum << endl;
        int cur_sum = 0, cur_len = 0, maxn = len; //当前连续段的和, 当前连续段的长度，最大连续段的长度
        bool suc = true; //能否这样划分
        for (int i = len + 1; i <= n; i++) {
            cur_sum += a[i], cur_len ++;
            // cout << "cur sum=" << cur_sum << endl;
            // cout << "cur len=" << cur_len << endl;
            if (cur_sum > sum) {
                suc = false;
                break;
            }
            if (cur_sum == sum) {
                maxn = max (cur_len, maxn);
                cur_len = cur_sum = 0;
            }
            //cout << "maxn=" << maxn << endl;
        }
        if (cur_sum && cur_sum < sum)   suc = false;
        //cout << "suc=" << suc << endl;
        if (suc)    ans = min (ans, maxn);
    }
    cout << ans << endl;
}

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

//划分成若干段，每一段的和都相等
//第一段是固定的，枚举第一段的长度，记录最大长度

D. Masha and a Beautiful Tree

每两个交换，然后每四个交换...依次按照两倍扩大

#include <bits/stdc++.h>

using namespace std;
const int N = 262144 + 5;
int a[N];

void solve () {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    if (is_sorted (a + 1, a + n + 1))   puts ("0");
    else {
        //逐层check
        long long ans = 0;
        for (int len = 1; len < n; len *= 2) {
            for (int i = 1; i <= n; i += len * 2) {
                int j = i + len;
                //cout << i << ' ' << j << endl;
                if (abs (a[j] - a[i]) != len) {
                    puts ("-1");
                    return ;
                }
                if (a[j] < a[i])    ans ++, swap (a[j], a[i]);
            }
            //for (int i = 1; i <= n; i++)    cout << a[i] << ' ';cout << endl;
        }
        cout << ans << endl;
    }
}

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

E. Sending a Sequence Over the Network

f[i]表示能转移到的状态，然后往左右跳a[i]，记录合法转移即可

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 5;
int a[N], f[N];

void solve () {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)    cin >> a[i], f[i] = 0;
    if (n == 1) {
        puts ("NO");
        return ;
    }
    f[0] = 1;
    for (int i = 1; i <= n; i++) {
        if (i + a[i] <= n)  f[i + a[i]] |= f[i - 1];
        if (i - a[i] >= 1)  f[i] |= f[i - 1 - a[i]];
    }
    if (f[n])   puts ("YES");
    else    puts ("NO");
}   

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

FG待补