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

void test_case() {
	int a, b, c, n; cin >> a >> b >> c >> n;
	c = min(c, n / 100); n -= c * 100;
	b = min(b, n / 10); n -= b * 10;
	a = min(a, n); n -= a;
	if (n == 0) cout << "YES\n";
	else cout << "NO\n";
}

int32_t main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int t; cin >> t;
	while (t--) test_case();
}

We only need to use $$$100$$$ coins as much as possible first, then $$$10$$$ coins as much as possible, and finally $$$1$$$ coins.

#ifndef LOCAL
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
#endif

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

#ifdef LOCAL
#include "../debug.h"
#else
#define dbg(...) "11-111"
#endif

int32_t main() {
    ios_base::sync_with_stdio(0);   cin.tie(0);

    int T;  cin >> T;
    while(T--) {
        int n;  cin >> n;
        vector<int> a(n), b(n);

        long long ans = LLONG_MIN, sum = 0, sa = 0, sb = 0;
        for(auto &i: a)     cin >> i;
        for(auto &i: b)     cin >> i, sum += i;


        ans = 2 * a.back() - sum;
        for(int i = n - 1 ; i >= 0 ; i--) {
            sa += a[i], sb += b[i];
            ans = max(ans, min((sa - a.back()) - (sb - b[i]), sa - (sum - b.back())));
        }

        cout << ans << '\n';

    }

    return 0;
}

Consider enumerate $$$i_a$$$ from $$$1$$$ to $$$n$$$.

For a fixed $$$i_a$$$,Bob has only two possible optimal options:choose $$$i_b=1$$$ or $$$i_b=i_a+1$$$ (if $$$i_a \neq n$$$).So Bob chooses $$$min(a_{i_a}+...+a_n-(b_1+...+b_{n-1}),a_{i_a}+...+a_{n-1}-(b_{i_a+1}+...+b_n))$$$.

Note $$$score_i$$$ as the answer for fixed $$$i_a=i$$$,the final answer is $$$max(score_i)$$$.

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

#define int long long

void test_case() {
	int n; cin >> n;
	string s; cin >> s;
	int same = 1, cnt = 0;
	for (int i = 1; i < n; i++) {
		if (s[i] != s[i - 1]) same = 0;
		same++;
		if (same >= 3) cnt += same - 2;
	}
	for (int i = 0; i < n; i += 2) s[i] = '1' + '0' - s[i];
	same = 1;
	for (int i = 1; i < n; i++) {
		if (s[i] != s[i - 1]) same = 0;
		same++;
		cnt += (same - 1) / 2;
	}
	cout << cnt << '\n';
}

int32_t main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int t; cin >> t;
	while (t--) test_case();
}

The key observation is that super palindromes are strings of length $$$\geq 3$$$ with all numbers equal, or a string with alternating $$$0-1$$$ (and with odd length).

After that, we only need to enumerate $$$i$$$ from $$$1$$$ to $$$n$$$ and count super palindromes that end in index $$$i$$$.

Coming soon...

Let $$$p$$$ be the lexicographically smallest balanced permutation of length $$$n$$$ and $$$x$$$ be the integer such that $$$1 \leq x < n$$$ and $$$p_1+p_2+⋯+p_x=p_{x+1}+p_{x+2}+⋯+p_n$$$. Let's take a look to the lexicographically smallest (not necessary balanced) permutation of length $$$n$$$, which is $$${1,2,\dots,n}$$$.

Let $$$m$$$ be the largest value such that $$$1+2+\dots+m \leq (m+1)+(m+2)+\dots+n$$$. Notice that since $$$m$$$ is as large as possible, $$$1+2+\dots+m+(m+1) > (m+2)+(m+3)+\dots+n$$$. Which means, $$$x < m+1$$$ for any balanced permutation of length $$$n$$$ (since no matter how you rearrange the permutation, $$$p_1+p_2+⋯+p_{m+1} \geq 1+2+\dots+m+(m+1)$$$). It is also worth noting that $$$m < n$$$. This will be useful later.

Since $$$p$$$ is the lexicographically smallest balanced permutation of length $$$n$$$, we want to greedily shove as much small numbers as possible to the prefix of $$$p$$$, then sort the prefix from smallest to largest. Notice that since $$$x < m+1$$$, it is possible that $$$x = m$$$ in the optimal solution and we can prove that it actually is.

Assume $$$p = {1,2,\dots,n}$$$ (in other words, $$$p_i = i$$$ for all $$$1 \leq i \leq n$$$). Focus on the $$$m$$$ first numbers in the permutation. Let $$$r = \frac{1+2+\dots+n}{2} - (1+2+\dots+m)$$$ (which means $$$r \geq 0$$$). Since $$$1+2+\dots+m+(m+1) > (m+2)+(m+3)+\dots+n$$$, $$$r < m+1$$$ which means $$$r \leq m$$$. This means that in order to make sure $$$p_1+p_2+⋯+p_m = p_{m+1}+p_{m+2}+⋯+p_n$$$, we can just add $$$1$$$ to all $$$p_j$$$ where $$$m-r+1 \leq j \leq m$$$ and $$$p_1,p_2,\dots,p_m$$$ will still be a possible prefix of the permutation, since $$$p_m \leq n$$$ and $$$p_1 < p_2 < \dots < p_m$$$ will still be satisfied after the changes (because $$$m < n$$$ and $$$p_1 < p_2 < \dots < p_{m-r} < p_{m-r+1} < \dots < p_m$$$). Now that we know $$$x = m$$$, how can we make $$$p$$$ as small lexicographically as possible?

Let's focus on the first $$$m$$$ numbers inside $$$p$$$. Set $$$p_i := i$$$ for all $$$1 \leq i \leq m$$$. We will modify this later. Now, since $$$r = \frac{1+2+\dots+n}{2} - (1+2+\dots+m)$$$, we need to add the value $$$r$$$ to the first $$$m$$$ numbers by spreading the addition. However, to make sure $$$p$$$ is as small as possible, we have to greedily add some value to some last numbers inside $$${p_1,p_2,\dots,p_m}$$$ such that the added value is $$$r$$$ and $$${p_1,p_2,\dots,p_m}$$$ is lexicographically smallest. In other words, we can do this procedure to construct the first $$$m$$$ numbers inside the permutation:

• Set $$$p_i := i$$$ for all $$$1 \leq i \leq m$$$.
• Let $$$j = m$$$. While $$$r > 0$$$, add $$$\min(r,k-j)$$$ to $$$p_j$$$, decrease $$$r$$$ by $$$\min(r,k-j)$$$, and decrease $$$j$$$ by $$$1$$$ where $$$k$$$ is the maximum integer such that $$$k \leq n$$$ and $$$k$$$ currently does not appear inside $$${p_1,p_2,\dots,p_m}$$$.

For the last $$$n-m$$$ numbers, notice that we can just fill these with all the numbers that are not used in the first $$$m$$$ numbers, since the sum of the numbers will automatically be equal to $$$p_1+p_2+\dots+p_m = \frac{1+2+\dots+n}{2}$$$. Do note that to make sure $$$p$$$ is as small as possible, $$$p_{m+1} < p_{m+2} < \dots < p_n$$$ need to be satisfied.

Bonus question: Can you guess which $$$n$$$ values results in no possible balanced permutation?

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

#define int long long
#define INF (int)(1e16)

#define N 510000
#define K 51

int a[N], cnt[N], pref[N], dp[K][N], n, k;

void calc(int i, int l, int r, int vl, int vr) {
	if (r < l) return;
	int mid = (l + r) / 2;
	int v, mn = INF;
	for (int j = vr; j >= vl; j--) {
		int sum = pref[mid] - pref[j - 1];
		int need = cnt[mid] - cnt[j - 1];
		if (i % 2 == 0) {
			need = (j + j + need - 1) * need / 2;
			sum = sum - need;
		} else {
			need = (mid + mid - need + 1) * need / 2;
			sum = need - sum;
		}
		if (dp[i - 1][j - 1] + sum < mn) {
			mn = dp[i - 1][j - 1] + sum;
			v = j;
		}
	}
	dp[i][mid] = mn;
	calc(i, l, mid - 1, vl, v);
	calc(i, mid + 1, r, v, vr);
}

void test_case() {
	cin >> n >> k;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for (int i = 2; i <= n; i++)
		a[i] ^= a[i - 1];
	for (int i = 1; i <= n; i++)
		cnt[i] = cnt[i - 1] + a[i], pref[i] = pref[i - 1] + i * a[i];
	for (int i = 1; i < n; i++)
		dp[0][i] = INF;
	dp[0][0] = 0;
	int ans = INF;
	for (int i = 1; i <= k; i++) {
		calc(i, 1, n - 1, 1, n - 1);
		if (i + (a[n] ^ (i & 1)) <= k)
			ans = min(ans, dp[i][n - 1]);
	}
	cout << ans << '\n';
}

int32_t main() {
	int t; cin >> t;
	while (t--) test_case();
}

Considering the prefix xor array of $$$a$$$, we found that the original operation was equivalent to swapping $$$prexor_ {i-1}, prexor_ {i} (1<i<n) $$$.

In fact,it is a variation of this.

Afterwards, the constraint of 'at most 2' is equivalent to having at most one continuous $$$1... 1$$$in $$$prexor$$$.

Because $$$prexor_n$$$ cannot be changed,if $$$prexor_n=1$$$, we merge all $$$1$$$ into one suffix. Otherwise, we will merge all $$$1$$$s into the vicinity of the middle $$$1$$$.

First,read solution(Easy Version).

In this version,we need to divide all $$$1$$$ s into up to $$$\lfloor \frac{k}{2} \rfloor$$$ groups and merge the $$$1$$$ s of each group in the optimal way.

Note $$$pos$$$ as positions of all $$$1$$$s,we can calculate the cost of merging $$$pos_l,...,pos_r$$$ in $$$O(1)$$$.Note it as $$$cost(l,r)$$$.

Now,we have a slmple $$$DP$$$:$$$dp_{i,j}=min_{0 \leq t <i}(dp_{t,j-1}+cost(t+1,i))$$$.It's $$$O(n^2k)$$$.How to optimize it?

We notice $$$cost(l+1,r+1)+cost(l,r) \leq cost(l+1,r)+cost(l,r+1)$$$ holds here.Due to the quadrilateral inequality, decision monotonicity holds. We can optimize it to $$$O(nk log n)$$$ using classic techniques (decision stack or divide and conquer).

The code above divides every group into the left and right parts to make calculations easier.

To learn the classic techniques:

CF 868 F

CF 321 E

#include <bits/stdc++.h>
using namespace std;
 
#define int long long
#define MOD (int)(1e9+7)
 
#define N 110000ll
#define K 20ll
#define B 850000006ll
 
struct segtree {
	vector<int> sum, lk, lb;
	segtree() {}
	segtree(int n) {
		sum = vector<int>(4 * n + 1, 0);
		lk = vector<int>(4 * n + 1, 1);
		lb = vector<int>(4 * n + 1, 0);
	}
	void push(int v, int l, int r) {
		if (lk[v] == 1 && lb[v] == 0) return;
		sum[v] = (lk[v] * sum[v] % MOD + (r - l + 1) * lb[v] % MOD) % MOD;
		if (l != r) {
			lk[v * 2 + 0] = lk[v] * lk[v * 2 + 0] % MOD;
			lb[v * 2 + 0] = (lk[v] * lb[v * 2 + 0] % MOD + lb[v]) % MOD;
			lk[v * 2 + 1] = lk[v] * lk[v * 2 + 1] % MOD;
			lb[v * 2 + 1] = (lk[v] * lb[v * 2 + 1] % MOD + lb[v]) % MOD;
		}
		lk[v] = 1, lb[v] = 0;
	}
	void upd(int v, int l, int r, int k, int x) {
		push(v, l, r);
		if (k < l || r < k) return;
		if (l == r) {
			sum[v] = x;
			return;
		}
		int mid = (l + r) / 2;
		upd(v * 2 + 0, l, mid, k, x);
		upd(v * 2 + 1, mid + 1, r, k, x);
		sum[v] = (sum[v * 2 + 0] + sum[v * 2 + 1]) % MOD;
	}
	void apply(int v, int l, int r, int ql, int qr, int k, int b) {
		push(v, l, r);
		if (qr < l || r < ql) return;
		if (ql <= l && r <= qr) {
			lk[v] = k;
			lb[v] = b;
			push(v, l, r);
			return;
		}
		int mid = (l + r) / 2;
		apply(v * 2 + 0, l, mid, ql, qr, k, b);
		apply(v * 2 + 1, mid + 1, r, ql, qr, k, b);
		sum[v] = (sum[v * 2 + 0] + sum[v * 2 + 1]) % MOD;
	}
	int get(int v, int l, int r, int ql, int qr) {
		push(v, l, r);
		if (qr < l || r < ql) return 0;
		if (ql <= l && r <= qr) return sum[v];
		int mid = (l + r) / 2;
		return (get(v * 2 + 0, l, mid, ql, qr) +
				get(v * 2 + 1, mid + 1, r, ql, qr)) % MOD;
	}
};
 
int a[N];
segtree p[K], pp;
 
int inv(int x) {
	return x == 1 ? 1 : MOD - MOD / x * inv(MOD % x) % MOD;
}
 
int32_t main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int n, q; cin >> n >> q;
	for (int i = 0; i < K; i++)
		p[i] = segtree(n);
	pp = segtree(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		for (int j = 0; j < K; j++)
			if (a[i] & (1 << j)) p[j].upd(1, 0, n - 1, i, 1);
	}
	int pk = (1 - 2 * B % MOD + MOD) % MOD, pb = B;
	int ppk = pk * pk % MOD, ppb = (B - B * B % MOD + MOD) % MOD;
	while (q--) {
		int t; cin >> t;
		if (t == 1) {
			int i, x; cin >> i >> x; i--;
			for (int j = 0; j < K; j++)
				p[j].upd(1, 0, n - 1, i, (x >> j) & 1);
			pp.upd(1, 0, n - 1, i, 0);
		} else if (t == 2) {
			int l, r; cin >> l >> r; l--; r--;
			for (int j = 0; j < K; j++)
				p[j].apply(1, 0, n - 1, l, r, pk, pb);
			pp.apply(1, 0, n - 1, l, r, ppk, ppb);
		} else if (t == 3) {
			int l, r; cin >> l >> r; l--; r--;
			if (l == r) {cout << 0 << '\n'; continue;}
			int ev = 0;
			int sumpp = pp.get(1, 0, n - 1, l, r);
			for (int j = 0; j < K; j++) {
				int sump = p[j].get(1, 0, n - 1, l, r);
				int cur = (sump * (r - l + 1 - sump + MOD) % MOD - sumpp + MOD) % MOD;
				(ev += cur * (1 << j) % MOD) %= MOD;
			}
			(ev *= inv(r - l + 1) * inv(r - l) % MOD * 2 % MOD) %= MOD;
			cout << ev << '\n';
		}
	}
}

Firstly, because of linearity of expectation, we can calculate answer for each bit independently.

Formula for expected value on segment for one bit is $$$\frac{\sum_{l\leq i\neq j\leq r} p_i(1-p_j)}{(r - l + 1) \cdot (r - l) / 2}$$$, where $$$p_i$$$ is probability of $$$a_i$$$ being set to $$$1$$$.

To calculate this, we just need to maintain $$$\sum_{l\leq i\leq r} p_i$$$ and $$$\sum_{l\leq i\leq r} p_i(1-p_i)$$$ Let $$$po_i$$$ be $$$p_i$$$ before second operation and $$$b$$$ probability of bit changing ($$$\frac{1}{20}$$$).

Now to maintain $$$\sum_{l\leq i\leq r} p_i$$$ and $$$\sum_{l\leq i\leq r} p_i(1-p_i)$$$ we just need to apply linear function on segment, we can do it using segment tree. This is already fast enough for C++20 compiler. We can notice that $$$p_i(1-p_i)$$$ is same for all bits, so we can maintain only one segtree for it. With this optimization solution passes using other compilers.

The final complexity is $$$\mathcal{O}(n \log n \log A)$$$