CodeForces Round #826 (Div.3) 康复训练

A

模拟题，不多说。

时间复杂度$O(3)$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>

const char ch[] = {'L', 'M', 'S'};
std::string s[2];
std::map<char, int> mp1, mp2;
int t;

void check() {
    for (int i = 0; i < 3; i++) {
        if (mp1[ch[i]] == 0 && mp2[ch[i]] == 0) continue;
        if (mp1[ch[i]] > 0 && mp2[ch[i]] == 0) {
            puts(">");
            return;
        }
        if (mp1[ch[i]] == 0 && mp2[ch[i]] > 0) {
            puts("<");
            return;
        }
        if (mp1[ch[i]] > 0 && mp2[ch[i]] > 0) {
            if (ch[i] == 'S') {
                if (mp1['X'] > mp2['X']) {
                    puts("<");
                    return;
                }
                if (mp1['X'] < mp2['X']) {
                    puts(">");
                    return;
                }
                if (mp1['X'] == mp2['X']) {
                    puts("=");
                    return;
                }
            }
            if (ch[i] == 'L') {
                if (mp1['X'] > mp2['X']) {
                    puts(">");
                    return;
                }
                if (mp1['X'] < mp2['X']) {
                    puts("<");
                    return;
                }
                if (mp1['X'] == mp2['X']) {
                    puts("=");
                    return;
                }
            }
            if (ch[i] == 'M') {
                puts("=");
                return;
            }
        }
    }
}

int main() {
    scanf("%d", &t);
    while (t--) {
        std::cin >> s[0] >> s[1];
        mp1.clear();
        mp2.clear();
        for (int i = 0; i < s[0].length(); i++) mp1[s[0][i]]++;
        for (int i = 0; i < s[1].length(); i++) mp2[s[1][i]]++;
        check();
    }
    return 0;
}

B

令$a_i=n-i+1$，若$a_i=i$,将其与其他位置交换。

还有更简单的构造，我比赛时瞎写了一堆，简直不能看。

时间复杂度$O(n)$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>

const int maxn = 2e5 + 7;
int t, n;
int a[maxn], b[maxn];
bool check;

int main() {
    scanf("%d", &t);
    while (t--) {
        check = 0;
        scanf("%d", &n);
        for (int i = n; i >= 1; i--) a[n - i + 1] = i;
        for (int i = 1; i <= n; i++) {
            if (a[i] == i && i + 2 != n) { std::swap(a[i + 1], a[i]); }
            if (a[i] == i && i + 2 == n) std::swap(a[i], a[i + 2]);
        }
        for (int i = 2; i < n; i++) {
            if (a[i - 1] != a[i] + 1 && a[i - 1] != a[i] - 1 && a[i + 1] != a[i] + 1 && a[i + 1] != a[i] - 1) check = 1;
        }
        if (a[1] != a[2] + 1 && a[1] != a[2] - 1) check = 1;
        if (a[n] != a[n - 1] + 1 && a[n] != a[n - 1] - 1) check = 1;
        if (check) puts("-1");
        if (!check) {
            for (int i = 1; i <= n; i++) std::cout << a[i] << " ";
            std::cout << std::endl;
        }
    }
    return 0;
}

C

GDOI2018 T1的弱化版。

满足条件的区间和一定是[1,n]区间和的因数。

考虑因数分解，每次分解判断是否有若干区间满足和相等，并更新最小的最大区间长。

其实一眼二分题，但我打暴力了。

时间复杂度$O(nlogn)$

#include<iostream>
#include<cstdio>
#include<cstring>

const int maxn = 2000 + 5;
int T;
int n;
int a[maxn], sum[maxn];
int Sum, cnt;
int dp[maxn];

int check2(int _sum) {
    int nowsum = 0;
    int maxlength = 0, i = 1;
    for (int j = 1; j <= n; j++) {
        nowsum += a[j];
        if (nowsum == _sum) {
            nowsum = 0;
            maxlength = std::max(maxlength, j - i + 1);
            i = j + 1;
        }
        if (nowsum > _sum) return -1;
    }
    return maxlength;
}

int check() {
    int minlength = n;
    for (int i = 1; i * i <= Sum; i++) {
        if (Sum % i != 0) continue;
        int len = check2(i);
        if (len != -1) minlength = std::min(minlength, len);
        int len2 = check2(Sum / i);
        if (len2 != -1) minlength = std::min(minlength, len2);
    }
    return minlength;
}

int main() {
    scanf("%d", &T);
    while (T--) {
        Sum = 0;
        cnt = 0;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]), Sum += a[i];
        std::cout << check() << std::endl;
    }
}

D

满足交换的条件一定是$a[l,mid].min=a[mid+1,r].max+1$

如果差额不为1，无论怎么交换左右子树都不能严格递增。

在push_up()里一边维护区间最大最小值，一边判断条件即可（不用真交换）。

时间复杂度$O(nlogn)$

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>

const int N = 3e5 + 7;
int a[N], ans;
int T, n;
bool check;
struct node {
    int l, r, Maxn, Minn;
} t[N << 2];

void push_up(int p) {
    bool flag=0;
    if (t[p << 1].Minn == t[p << 1 | 1].Maxn + 1) ans++,flag=1;// 45 23
    if (t[p << 1].Maxn == t[p << 1 | 1].Minn - 1) flag=1;// 23 45
    if(!flag) check = 1;//无法构造beautiful tree
    t[p].Maxn = std::max(t[p << 1].Maxn, t[p << 1 | 1].Maxn);
    t[p].Minn = std::min(t[p << 1].Minn, t[p << 1 | 1].Minn);
}

void build(int p, int l, int r) {
    t[p].l = l;
    t[p].r = r;
    if (l == r) {
        t[p].Maxn = t[p].Minn = a[l];
        return;
    }
    int mid = l + r >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    push_up(p);
}

int main() {
    scanf("%d", &T);
    while (T--) {
        check = 0;
        ans = 0;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        build(1, 1, n);
        if (check) puts("-1");
        else std::cout << ans << std::endl;
    }
}

E（赛后）

自己的做法没过，参考的@cup-pyy。

设$f[i]$为$b$中前$i$个元素是否能够还原成$a$

$b[i+1]$可以是右侧子序列的长度，所以利用这点可以推出：

①当$i$合法时，$i+b[i+1]+1$是合法的

②且当$i-b[i]-1$是合法的时候（$b[i]$为左侧子序列的长度），$i$是合法的

需要注意：要先判断第②点，不然会漏掉$b[i+b[i+1]+1]$的确认。

#include<iostream>
#include<cstdio>
#include<cstring>

const int maxn = 2e5 + 7;
int f[maxn];
int T, n, b[maxn];

int main() {
    scanf("%d", &T);
    while (T--) {
        memset(f, 0, sizeof(f));
        f[0] = 1;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
        for (int i = 0; i <= n; i++) {
            if (i - b[i] - 1 >= 0 && f[i - b[i] - 1]) f[i] = 1;
            if (f[i] && i + b[i + 1] + 1 <= n) f[i + b[i + 1] + 1] = 1;
        }
        puts(f[n] ? "YES" : "NO");
    }
}

剩下的没做，看什么时候有空再做吧...