A. Split the Multiset
Problem Statement
A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, { 2 , 2 , 4 } \{2,2,4\} {2,2,4} is a multiset.
You have a multiset S S S. Initially, the multiset contains only one positive integer n n n. That is, S = { n } S=\{n\} S={n}. Additionally, there is a given positive integer k k k.
In one operation, you can select any positive integer u u u in S S S and remove one copy of u u u from S S S. Then, insert no more than k k k positive integers into S S S so that the sum of all inserted integers is equal to u u u.
Find the minimum number of operations to make S S S contain n n n ones.
Input
Each test contains multiple test cases. The first line contains the number of test cases t t t ( 1 ≤ t ≤ 1000 1 \le t \le 1000 1≤t≤1000). Description of the test cases follows.
The only line of each testcase contains two integers n , k n,k n,k ( 1 ≤ n ≤ 1000 , 2 ≤ k ≤ 1000 1\le n\le 1000,2\le k\le 1000 1≤n≤1000,2≤k≤1000).
Output
For each testcase, print one integer, which is the required answer.
Example
| input |
|---|
| 4 |
| 1 5 |
| 5 2 |
| 6 3 |
| 16 4 |
| output |
|---|
| 0 |
| 4 |
| 3 |
| 5 |
Note
For the first test case, initially S = { 1 } S=\{1\} S={1}, already satisfying the requirement. Therefore, we need zero operations.
For the second test case, initially S = { 5 } S=\{5\} S={5}. We can apply the following operations:
- Select u = 5 u=5 u=5, remove u u u from S S S, and insert 2 , 3 2,3 2,3 into S S S. Now, S = { 2 , 3 } S=\{2,3\} S={2,3}.
- Select u = 2 u=2 u=2, remove u u u from S S S, and insert 1 , 1 1,1 1,1 into S S S. Now, S = { 1 , 1 , 3 } S=\{1,1,3\} S={1,1,3}.
- Select u = 3 u=3 u=3, remove u u u from S S S, and insert 1 , 2 1,2 1,2 into S S S. Now, S = { 1 , 1 , 1 , 2 } S=\{1,1,1,2\} S={1,1,1,2}.
- Select u = 2 u=2 u=2, remove u u u from S S S, and insert 1 , 1 1,1 1,1 into S S S. Now, S = { 1 , 1 , 1 , 1 , 1 } S=\{1,1,1,1,1\} S={1,1,1,1,1}.
Using 4 4 4 operations in total, we achieve the goal.
For the third test case, initially S = { 6 } S=\{6\} S={6}. We can apply the following operations:
- Select u = 6 u=6 u=6, remove u u u from S S S, and insert 1 , 2 , 3 1,2,3 1,2,3 into S S S. Now, S = { 1 , 2 , 3 } S=\{1,2,3\} S={1,2,3}.
- Select u = 2 u=2 u=2, remove u u u from S S S, and insert 1 , 1 1,1 1,1 into S S S. Now, S = { 1 , 1 , 1 , 3 } S=\{1,1,1,3\} S={1,1,1,3}.
- Select u = 3 u=3 u=3, remove u u u from S S S, and insert 1 , 1 , 1 1,1,1 1,1,1 into S S S. Now, S = { 1 , 1 , 1 , 1 , 1 , 1 } S=\{1,1,1,1,1,1\} S={1,1,1,1,1,1}.
Using 3 3 3 operations in total, we achieve the goal.
For the fourth test case, initially S = { 16 } S=\{16\} S={16}. We can apply the following operations:
- Select u = 16 u=16 u=16, remove u u u from S S S, and insert 4 , 4 , 4 , 4 4,4,4,4 4,4,4,4 into S S S. Now, S = { 4 , 4 , 4 , 4 } S=\{4,4,4,4\} S={4,4,4,4}.
- Repeat for 4 4 4 times: select u = 4 u=4 u=4, remove u u u from S S S, and insert 1 , 1 , 1 , 1 1,1,1,1 1,1,1,1 into S S S.
Using 5 5 5 operations in total, we achieve the goal.
Solution
具体见文后视频。
Code
cpp
#include <bits/stdc++.h>
void solve() {
int n, k;
std::cin >> n >> k;
std::cout << (n + k - 3) / (k - 1) << "\n";
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int t;
std::cin >> t;
while (t--) {
solve();
}
return 0;
}B. Make Majority
Problem Statement
You are given a sequence [ a 1 , ... , a n ] [a_1,\ldots,a_n] [a1,...,an] where each element a i a_i ai is either 0 0 0 or 1 1 1. You can apply several (possibly zero) operations to the sequence. In each operation, you select two integers 1 ≤ l ≤ r ≤ ∣ a ∣ 1\le l\le r\le |a| 1≤l≤r≤∣a∣ (where ∣ a ∣ |a| ∣a∣ is the current length of a a a) and replace [ a l , ... , a r ] [a_l,\ldots,a_r] [al,...,ar] with a single element x x x, where x x x is the majority of [ a l , ... , a r ] [a_l,\ldots,a_r] [al,...,ar].
Here, the majority of a sequence consisting of 0 0 0 and 1 1 1 is defined as follows: suppose there are c 0 c_0 c0 zeros and c 1 c_1 c1 ones in the sequence, respectively.
- If c 0 ≥ c 1 c_0\ge c_1 c0≥c1, the majority is 0 0 0.
- If c 0 < c 1 c_0<c_1 c0<c1, the majority is 1 1 1.
For example, suppose a = [ 1 , 0 , 0 , 0 , 1 , 1 ] a=[1,0,0,0,1,1] a=[1,0,0,0,1,1]. If we select l = 1 , r = 2 l=1,r=2 l=1,r=2, the resulting sequence will be [ 0 , 0 , 0 , 1 , 1 ] [0,0,0,1,1] [0,0,0,1,1]. If we select l = 4 , r = 6 l=4,r=6 l=4,r=6, the resulting sequence will be [ 1 , 0 , 0 , 1 ] [1,0,0,1] [1,0,0,1].
Determine if you can make a = [ 1 ] a=[1] a=[1] with a finite number of operations.
Input
Each test contains multiple test cases. The first line contains the number of test cases t t t ( 1 ≤ t ≤ 4 ⋅ 1 0 4 1 \le t \le 4\cdot 10^4 1≤t≤4⋅104). Description of the test cases follows.
The first line of each testcase contains one integer n n n ( 1 ≤ n ≤ 2 ⋅ 1 0 5 1\le n\le 2\cdot 10^5 1≤n≤2⋅105).
The second line of each testcase contains a string consisting of 0 0 0 and 1 1 1, describing the sequence a a a.
It's guaranteed that the sum of n n n over all testcases does not exceed 2 ⋅ 1 0 5 2\cdot 10^5 2⋅105.
Output
For each testcase, if it's possible to make a = [ 1 ] a=[1] a=[1], print YES. Otherwise, print NO. You can output the answer in any case (upper or lower). For example, the strings yEs, yes, Yes, and YES will be recognized as positive responses.
Example
| input |
|---|
| 5 |
| 1 |
| 0 |
| 1 |
| 1 |
| 2 |
| 01 |
| 9 |
| 100000001 |
| 9 |
| 000011000 |
| output |
|---|
| No |
| Yes |
| No |
| Yes |
| No |
Note
In the fourth testcase of the example, initially a = [ 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 ] a=[1,0,0,0,0,0,0,0,1] a=[1,0,0,0,0,0,0,0,1]. A valid sequence of operations is:
- Select l = 2 , r = 8 l=2,r=8 l=2,r=8 and apply the operation. a a a becomes [ 1 , 0 , 1 ] [1,0,1] [1,0,1].
- Select l = 1 , r = 3 l=1,r=3 l=1,r=3 and apply the operation. a a a becomes [ 1 ] [1] [1].
Solution
具体见文后视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
void solve() {
int n;
string s;
cin >> n >> s;
int one = 0, zero = 0, len = 0;
for (int i = 0; i < s.size(); i ++)
if (s[i] == '0') len ++;
else {
if (len) zero ++;
one ++, len = 0;
}
if (len) zero ++;
if (one > zero) cout << "Yes" << endl;
else cout << "No" << endl;
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}C. Increasing Sequence with Fixed OR
Problem Statement
You are given a positive integer n n n. Find the longest sequence of positive integers a = [ a 1 , a 2 , ... , a k ] a=[a_1,a_2,\ldots,a_k] a=[a1,a2,...,ak] that satisfies the following conditions, and print the sequence:
- a i ≤ n a_i\le n ai≤n for all 1 ≤ i ≤ k 1\le i\le k 1≤i≤k.
- a a a is strictly increasing. That is, a i > a i − 1 a_i>a_{i-1} ai>ai−1 for all 2 ≤ i ≤ k 2\le i\le k 2≤i≤k.
- a i ∣ a i − 1 = n a_i\,|\,a_{i-1}=n ai∣ai−1=n for all 2 ≤ i ≤ k 2\le i\le k 2≤i≤k, where ∣ | ∣ denotes the bitwise OR operation.
Input
Each test contains multiple test cases. The first line contains the number of test cases t t t ( 1 ≤ t ≤ 1000 1 \le t \le 1000 1≤t≤1000). Description of the test cases follows.
The only line of each test case contains one integer n n n ( 1 ≤ n ≤ 1 0 18 1\le n\le 10^{18} 1≤n≤1018).
It's guaranteed that the sum of lengths of the longest valid sequences does not exceed 5 ⋅ 1 0 5 5\cdot 10^5 5⋅105.
Output
For each testcase, print two lines. In the first line, print the length of your constructed sequence, k k k. In the second line, print k k k positive integers, denoting the sequence. If there are multiple longest sequences, you can print any of them.
Example
| input |
|---|
| 4 |
| 1 |
| 3 |
| 14 |
| 23 |
| output |
|---|
| 1 |
| 1 |
| 3 |
| 1 2 3 |
| 4 |
| 4 10 12 14 |
| 5 |
| 7 18 21 22 23 |
Solution
具体见文后视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
void solve() {
int n;
cin >> n;
std::vector<int> pos;
for (int i = 0; i < 62; i ++)
if (n >> i & 1) pos.push_back(i);
int m = pos.size();
std::vector<int> way;
for (int i = m - 1; i >= 0; i --) {
int tmp = 0;
for (int j = m - 1; j > i; j --)
tmp |= (1ll << j);
for (int j = i - 1, k = 1; j >= 0; j --)
tmp |= (k << j), k ^= 1;
if (tmp) way.push_back(tmp);
}
way.push_back((1ll << m) - 1);
std::vector<int> res;
for (auto v : way) {
int tmp = 0;
for (int i = 0; i < pos.size(); i ++)
tmp += ((v >> i & 1) << pos[i]);
res.push_back(tmp);
}
cout << res.size() << endl;
for (auto v : res)
cout << v << " ";
cout << endl;
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}D. The Omnipotent Monster Killer
Problem Statement
You, the monster killer, want to kill a group of monsters. The monsters are on a tree with n n n vertices. On vertex with number i i i ( 1 ≤ i ≤ n 1\le i\le n 1≤i≤n), there is a monster with a i a_i ai attack points. You want to battle with monsters for 1 0 100 10^{100} 10100 rounds.
In each round, the following happens in order:
- All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.
- You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.
There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.
If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?
Input
Each test contains multiple test cases. The first line contains the number of test cases t t t ( 1 ≤ t ≤ 1 0 4 1 \le t \le 10^4 1≤t≤104). Description of the test cases follows.
The first line of each test case contains an integer n n n ( 1 ≤ n ≤ 3 ⋅ 1 0 5 1\le n\le 3\cdot 10^5 1≤n≤3⋅105).
The second line of each test case contains n n n integers a 1 , ... , a n a_1,\ldots,a_n a1,...,an ( 1 ≤ a i ≤ 1 0 12 1\le a_i\le 10^{12} 1≤ai≤1012).
The following n − 1 n-1 n−1 lines each contain two integers x , y x,y x,y ( 1 ≤ x , y ≤ n 1\le x,y\le n 1≤x,y≤n), denoting an edge on the tree connecting vertex x x x and y y y.
It is guaranteed that the sum of n n n over all test cases does not exceed 3 ⋅ 1 0 5 3\cdot 10^5 3⋅105.
Output
For each test case, print one integer: the minimum possible health decrement.
Example
| input |
|---|
| 3 |
| 1 |
| 1000000000000 |
| 5 |
| 47 15 32 29 23 |
| 1 2 |
| 1 3 |
| 2 4 |
| 2 5 |
| 7 |
| 8 10 2 3 5 7 4 |
| 1 2 |
| 1 4 |
| 3 2 |
| 5 3 |
| 6 2 |
| 7 5 |
| output |
|---|
| 1000000000000 |
| 193 |
| 57 |
Note
In the first test case, an optimal sequence of operations would be:
- In the first round: first, receive the attack from the monster on vertex 1 1 1, so your health decreases by 1 0 12 10^{12} 1012. Then kill the monster on vertex 1 1 1.
- In the second round to the 1 0 100 10^{100} 10100-th round: all monsters have been killed, so nothing happens.
The total health decrement is 1 0 12 10^{12} 1012.
In the second test case, an optimal sequence of operations would be:
- In the first round: first, receive the attack from the monster on vertex 1 , 2 , 3 , 4 , 5 1,2,3,4,5 1,2,3,4,5, so your health decreases by 47 + 15 + 32 + 29 + 23 = 146 47+15+32+29+23=146 47+15+32+29+23=146. Then kill the monsters on vertex 1 , 4 , 5 1,4,5 1,4,5.
- In the second round: first, receive the attack from the monster on vertex 2 , 3 2,3 2,3, so your health decreases by 15 + 32 = 47 15+32=47 15+32=47. Then kill the monsters on vertex 2 , 3 2,3 2,3.
- In the third round to the 1 0 100 10^{100} 10100-th round: all monsters have been killed, so nothing happens.
The total health decrement is 193 193 193.
In the third test case, an optimal sequence of operations would be:
- In the first round: first, receive the attack from the monster on vertex 1 , 2 , 3 , 4 , 5 , 6 , 7 1,2,3,4,5,6,7 1,2,3,4,5,6,7, so your health decreases by 8 + 10 + 2 + 3 + 5 + 7 + 4 = 39 8+10+2+3+5+7+4=39 8+10+2+3+5+7+4=39. Then kill the monsters on vertex 1 , 3 , 6 , 7 1,3,6,7 1,3,6,7.
- In the second round: first, receive the attack from the monster on vertex 2 , 4 , 5 2,4,5 2,4,5, so your health decreases by 10 + 3 + 5 = 18 10+3+5=18 10+3+5=18. Then kill the monsters on vertex 2 , 4 , 5 2,4,5 2,4,5.
- In the third round to the 1 0 100 10^{100} 10100-th round: all monsters have been killed, so nothing happens.
The total health decrement is 57 57 57.
Solution
具体见文后视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 3e5 + 10;
int n;
int a[N], dp[N][30];
std::vector<int> g[N];
void dfs(int u, int fa) {
for (int i = 1; i <= 25; i ++) dp[u][i] = a[u] * i;
for (auto v : g[u]) {
if (v == fa) continue;
dfs(v, u);
for (int i = 1; i <= 25; i ++) {
int mn = 9e18;
for (int j = 1; j <= 25; j ++)
if (i != j) mn = min(mn, dp[v][j]);
dp[u][i] += mn;
}
}
}
void solve() {
cin >> n;
int tot = 0;
for (int i = 1; i <= n; i ++) cin >> a[i];
for (int i = 1; i < n; i ++) {
int u, v;
cin >> u >> v;
g[u].push_back(v), g[v].push_back(u);
}
dfs(1, -1);
int res = 9e18;
for (int i = 1; i <= 25; i ++) res = min(res, dp[1][i]);
cout << res << endl;
for (int i = 1; i <= n; i ++) g[i].clear();
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}E. Range Minimum Sum
Problem Statement
For an array [ a 1 , a 2 , ... , a n ] [a_1,a_2,\ldots,a_n] [a1,a2,...,an] of length n n n, define f ( a ) f(a) f(a) as the sum of the minimum element over all subsegments. That is,
f ( a ) = ∑ l = 1 n ∑ r = l n min l ≤ i ≤ r a i . f(a)=\sum_{l=1}^n\sum_{r=l}^n \min_{l\le i\le r}a_i. f(a)=l=1∑nr=l∑nl≤i≤rminai.
A permutation is a sequence of integers from 1 1 1 to n n n of length n n n containing each number exactly once. You are given a permutation [ a 1 , a 2 , ... , a n ] [a_1,a_2,\ldots,a_n] [a1,a2,...,an]. For each i i i, solve the following problem independently:
- Erase a i a_i ai from a a a, concatenating the remaining parts, resulting in b = [ a 1 , a 2 , ... , a i − 1 , a i + 1 , ... , a n ] b = [a_1,a_2,\ldots,a_{i-1},\;a_{i+1},\ldots,a_{n}] b=[a1,a2,...,ai−1,ai+1,...,an].
- Calculate f ( b ) f(b) f(b).
Input
Each test contains multiple test cases. The first line contains the number of test cases t t t ( 1 ≤ t ≤ 1 0 5 1 \le t \le 10^5 1≤t≤105). Description of the test cases follows.
The first line of each test case contains an integer n n n ( 1 ≤ n ≤ 5 ⋅ 1 0 5 1\le n\le 5\cdot 10^5 1≤n≤5⋅105).
The second line of each test case contains n n n distinct integers a 1 , ... , a n a_1,\ldots,a_n a1,...,an ( 1 ≤ a i ≤ n 1\le a_i\le n 1≤ai≤n).
It is guaranteed that the sum of n n n over all test cases does not exceed 1 0 6 10^6 106.
Output
For each test case, print one line containing n n n integers. The i i i-th integer should be the answer when erasing a i a_i ai.
Example
| input |
|---|
| 4 |
| 1 |
| 1 |
| 3 |
| 3 1 2 |
| 5 |
| 4 2 1 5 3 |
| 8 |
| 8 1 4 6 7 3 5 2 |
| output |
|---|
| 0 |
| 4 7 5 |
| 19 21 27 17 19 |
| 79 100 72 68 67 80 73 80 |
Note
In the second test case, a = [ 3 , 1 , 2 ] a=[3,1,2] a=[3,1,2].
- When removing a 1 a_1 a1, b = [ 1 , 2 ] b=[1,2] b=[1,2]. f ( b ) = 1 + 2 + min { 1 , 2 } = 4 f(b)=1+2+\min\{1,2\}=4 f(b)=1+2+min{1,2}=4.
- When removing a 2 a_2 a2, b = [ 3 , 2 ] b=[3,2] b=[3,2]. f ( b ) = 3 + 2 + min { 3 , 2 } = 7 f(b)=3+2+\min\{3,2\}=7 f(b)=3+2+min{3,2}=7.
- When removing a 3 a_3 a3, b = [ 3 , 1 ] b=[3,1] b=[3,1]. f ( b ) = 3 + 1 + min { 3 , 1 } = 5 f(b)=3+1+\min\{3,1\}=5 f(b)=3+1+min{3,1}=5.
Solution
具体见文后视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 5e5 + 10;
int n;
int a[N], l[N], r[N], dcr[N];
int f[N][22];
void build() {
for (int j = 0; j <= log2(n); j ++)
for (int i = 1; i + (1ll << j) - 1 <= n; i ++)
if (!j) f[i][j] = a[i];
else f[i][j] = min(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
}
int query(int l, int r) {
int k = log2(r - l + 1);
return min(f[l][k], f[r - (1ll << k) + 1][k]);
}
void solve() {
cin >> n;
for (int i = 1; i <= n; i ++) cin >> a[i];
stack<int> stk;
for (int i = 1; i <= n; i ++) {
while (stk.size() && a[stk.top()] >= a[i]) stk.pop();
if (stk.size()) l[i] = i - stk.top();
else l[i] = i;
stk.push(i);
}
while (stk.size()) stk.pop();
for (int i = n; i >= 1; i --) {
while (stk.size() && a[stk.top()] >= a[i]) stk.pop();
if (stk.size()) r[i] = stk.top() - i;
else r[i] = n - i + 1;
stk.push(i);
}
build();
int tot = 0;
for (int i = 1; i <= n; i ++) tot += a[i] * l[i] * r[i];
for (int i = 1; i <= n; i ++) dcr[i] = 0;
for (int i = 1; i <= n; i ++) dcr[i - l[i] + 1] += r[i] * a[i], dcr[i] -= r[i] * a[i];
for (int i = 1; i <= n; i ++) dcr[i + 1] += l[i] * a[i], dcr[i + r[i]] -= l[i] * a[i];
for (int i = 1; i <= n; i ++) dcr[i] += dcr[i - 1];
for (int i = 1; i <= n; i ++) {
if (i - l[i] > 1) {
int L = 1, R = i - l[i] - 1, res = i - l[i];
while (L <= R) {
int mid = L + R >> 1;
if (query(mid, i - l[i] - 1) > a[i]) R = mid - 1, res = min(res, mid);
else L = mid + 1;
}
dcr[i - l[i]] -= (i - l[i] - res) * a[i] * r[i];
}
if (i + r[i] < n) {
int L = i + r[i] + 1, R = n, res = i + r[i];
while (L <= R) {
int mid = L + R >> 1;
if (query(i + r[i] + 1, mid) > a[i]) L = mid + 1, res = max(res, mid);
else R = mid - 1;
}
dcr[i + r[i]] -= (res - i - r[i]) * a[i] * l[i];
}
}
for (int i = 1; i <= n; i ++)
cout << tot - a[i] * l[i] * r[i] - dcr[i] << " ";
cout << endl;
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}视频讲解
Codeforces Round 958 (Div. 2)（A ~ E 讲解）
最后祝大家早日