A. Sliding
Problem Statement
Red was ejected. They were not the imposter.
There are n n n rows of m m m people. Let the position in the r r r-th row and the c c c-th column be denoted by ( r , c ) (r, c) (r,c). Number each person starting from 1 1 1 in row-major order, i.e., the person numbered ( r − 1 ) ⋅ m + c (r-1)\cdot m+c (r−1)⋅m+c is initially at ( r , c ) (r,c) (r,c).
The person at ( r , c ) (r, c) (r,c) decides to leave. To fill the gap, let the person who left be numbered i i i. Each person numbered j > i j>i j>i will move to the position where the person numbered j − 1 j-1 j−1 is initially at. The following diagram illustrates the case where n = 2 n=2 n=2, m = 3 m=3 m=3, r = 1 r=1 r=1, and c = 2 c=2 c=2.
Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at ( r 0 , c 0 ) (r_0, c_0) (r0,c0) and then moved to ( r 1 , c 1 ) (r_1, c_1) (r1,c1), the Manhattan distance is ∣ r 0 − r 1 ∣ + ∣ c 0 − c 1 ∣ |r_0-r_1|+|c_0-c_1| ∣r0−r1∣+∣c0−c1∣.
Input
The first line contains a single integer t t t ( 1 ≤ t ≤ 1 0 4 1\le t\le 10^4 1≤t≤104) --- the number of test cases.
The only line of each testcase contains 4 4 4 integers n n n, m m m, r r r, and c c c ( 1 ≤ r ≤ n ≤ 1 0 6 1\le r\le n\le 10^6 1≤r≤n≤106, 1 ≤ c ≤ m ≤ 1 0 6 1 \le c \le m \le 10^6 1≤c≤m≤106), where n n n is the number of rows, m m m is the number of columns, and ( r , c ) (r,c) (r,c) is the position where the person who left is initially at.
Output
For each test case, output a single integer denoting the sum of the Manhattan distances.
Example
| input |
|---|
| 4 |
| 2 3 1 2 |
| 2 2 2 1 |
| 1 1 1 1 |
| 1000000 1000000 1 1 |
| output |
|---|
| 6 |
| 1 |
| 0 |
| 1999998000000 |
note
For the first test case, the person numbered 2 2 2 leaves, and the distances of the movements of the person numbered 3 3 3, 4 4 4, 5 5 5, and 6 6 6 are 1 1 1, 3 3 3, 1 1 1, and 1 1 1, respectively. So the answer is 1 + 3 + 1 + 1 = 6 1+3+1+1=6 1+3+1+1=6.
For the second test case, the person numbered 3 3 3 leaves, and the person numbered 4 4 4 moves. The answer is 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, m, r, c;
cin >> n >> m >> r >> c;
int res = m - c;
res += m * (n - r);
res += (n - r) * (m - 1);
cout << res << endl;
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}B. Everyone Loves Tres
Problem Statement
There are 3 heroes and 3 villains, so 6 people in total.
Given a positive integer n n n. Find the smallest integer whose decimal representation has length n n n and consists only of 3 3 3s and 6 6 6s such that it is divisible by both 33 33 33 and 66 66 66. If no such integer exists, print − 1 -1 −1.
Input
The first line contains a single integer t t t ( 1 ≤ t ≤ 500 1\le t\le 500 1≤t≤500) --- the number of test cases.
The only line of each test case contains a single integer n n n ( 1 ≤ n ≤ 500 1\le n\le 500 1≤n≤500) --- the length of the decimal representation.
Output
For each test case, output the smallest required integer if such an integer exists and − 1 -1 −1 otherwise.
Example
| input |
|---|
| 6 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 7 |
| output |
|---|
| -1 |
| 66 |
| -1 |
| 3366 |
| 36366 |
| 3336366 |
note
For n = 1 n=1 n=1, no such integer exists as neither 3 3 3 nor 6 6 6 is divisible by 33 33 33.
For n = 2 n=2 n=2, 66 66 66 consists only of 6 6 6s and it is divisible by both 33 33 33 and 66 66 66.
For n = 3 n=3 n=3, no such integer exists. Only 363 363 363 is divisible by 33 33 33, but it is not divisible by 66 66 66.
For n = 4 n=4 n=4, 3366 3366 3366 and 6666 6666 6666 are divisible by both 33 33 33 and 66 66 66, and 3366 3366 3366 is the smallest.
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;
if (n == 1 || n == 3) {
cout << -1 << endl;
return;
}
if (n % 2 == 0) {
for (int i = 1; i <= n - 2; i ++)
cout << 3;
cout << 66 << endl;
} else {
for (int i = 1; i <= n - 5; i ++)
cout << 3;
cout << 36366 << endl;
}
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}C. Alya and Permutation
Problem Statement
Alya has been given a hard problem. Unfortunately, she is too busy running for student council. Please solve this problem for her.
Given an integer n n n, construct a permutation p p p of integers 1 , 2 , ... , n 1, 2, \ldots, n 1,2,...,n that maximizes the value of k k k (which is initially 0 0 0) after the following process.
Perform n n n operations, on the i i i-th operation ( i = 1 , 2 , ... , n i=1, 2, \dots, n i=1,2,...,n),
- If i i i is odd, k = k & p i k=k\,\&\,p_i k=k&pi, where & \& & denotes the bitwise AND operation.
- If i i i is even, k = k ∣ p i k=k\,|\,p_i k=k∣pi, where ∣ | ∣ denotes the bitwise OR operation.
Input
The first line contains a single integer t t t ( 1 ≤ t ≤ 500 1\le t\le 500 1≤t≤500) --- the number of test cases.
The only line of each test case contains a single integer n n n ( 5 ≤ n ≤ 2 ⋅ 1 0 5 5\le n\le 2 \cdot 10^5 5≤n≤2⋅105) --- the length of the permutation.
It is guaranteed that the sum of n n n over all test cases does not exceed 2 ⋅ 1 0 5 2 \cdot 10^5 2⋅105.
Output
For each test case, output the maximum value of k k k in the first line and output the permutation p 1 , p 2 , ... , p n p_1, p_2,\ldots, p_n p1,p2,...,pn in the second line.
If there are multiple such permutations, output any.
Example
| input |
|---|
| 6 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
| 10 |
| output |
|---|
| 5 |
| 2 1 3 4 5 |
| 7 |
| 1 2 4 6 5 3 |
| 7 |
| 2 4 5 1 3 6 7 |
| 15 |
| 2 4 5 1 3 6 7 8 |
| 9 |
| 2 4 5 6 7 1 3 8 9 |
| 15 |
| 1 2 3 4 5 6 8 10 9 7 |
note
For the first test case, the value of k k k is determined as follows:
k = 0 k = 0 k=0 initially.
- On the 1 1 1st operation, 1 1 1 is odd, so Alya sets k k k to be k & p 1 = 0 & 2 = 0 k\&p_1 = 0\&2 = 0 k&p1=0&2=0.
- On the 2 2 2nd operation, 2 2 2 is even, so Alya sets k k k to be k ∣ p 2 = 0 ∣ 1 = 1 k|p_2 = 0|1 = 1 k∣p2=0∣1=1.
- On the 3 3 3rd operation, 3 3 3 is odd, so Alya sets k k k to be k & p 3 = 1 & 3 = 1 k\&p_3 = 1\&3 = 1 k&p3=1&3=1.
- On the 4 4 4th operation, 4 4 4 is even, so Alya sets k k k to be k ∣ p 4 = 1 ∣ 4 = 5 k|p_4 = 1|4 = 5 k∣p4=1∣4=5.
- On the 5 5 5th operation, 5 5 5 is odd, so Alya sets k k k to be k & p 5 = 5 & 5 = 5 k\&p_5 = 5\&5 = 5 k&p5=5&5=5.
The final value of k k k is 5 5 5. It can be shown that the final value of k k k is at most 5 5 5 for all permutations of length 5 5 5. Another valid output is [ 2 , 3 , 1 , 4 , 5 ] [2, 3, 1, 4, 5] [2,3,1,4,5].
For the second test case, the final value of k k k is 7 7 7. It can be shown that the final value of k k k is at most 7 7 7 for all permutations of length 6 6 6. Other valid outputs include [ 2 , 4 , 1 , 6 , 3 , 5 ] [2, 4, 1, 6, 3, 5] [2,4,1,6,3,5] and [ 5 , 2 , 6 , 1 , 3 , 4 ] [5, 2, 6, 1, 3, 4] [5,2,6,1,3,4].
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> res;
int lst = -1;
if (n & 1) lst = n --;
int sep = 1;
while (sep * 2 <= n) sep *= 2;
sep --;
for (int i = 1; i <= n; i ++)
if (i != 1 && i != 3 && i != sep - 1 && i != sep && i != n)
res.push_back(i);
std::vector<int> final = {1, 3, sep - 1, sep, n};
if (sep == 3) final = {2, 1, 3, n};
if (~lst) final.push_back(lst);
for (auto v : final) res.push_back(v);
if (~lst) cout << lst << endl;
else cout << sep * 2 + 1 << 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. Yet Another Real Number Problem
Problem Statement
Your task is to maximize the sum of the array after performing any number of such operations. Since it could be large, output this sum modulo 109+7
.Three r there are's in strawberry.
You are given an array b b b of length m m m. You can perform the following operation any number of times (possibly zero):
- Choose two distinct indices i i i and j j j where 1 ≤ i < j ≤ m \bf{1\le i < j\le m} 1≤i<j≤m and b i b_i bi is even, divide b i b_i bi by 2 2 2 and multiply b j b_j bj by 2 2 2.
Your task is to maximize the sum of the array after performing any number of such operations. Since it could be large, output this sum modulo 1 0 9 + 7 10^9+7 109+7.
Since this problem is too easy, you are given an array a a a of length n n n and need to solve the problem for each prefix of a a a.
In other words, denoting the maximum sum of b b b after performing any number of such operations as f ( b ) f(b) f(b), you need to output f ( [ a 1 ] ) f([a_1]) f([a1]), f ( [ a 1 , a 2 ] ) f([a_1,a_2]) f([a1,a2]), ... \ldots ..., f ( [ a 1 , a 2 , ... , a n ] ) f([a_1,a_2,\ldots,a_n]) f([a1,a2,...,an]) modulo 1 0 9 + 7 10^9+7 109+7 respectively.
Input
The first line contains a single integer t t t ( 1 ≤ t ≤ 1 0 4 1\le t\le 10^4 1≤t≤104) --- the number of test cases.
The first line of each test case contains a single integer n n n ( 1 ≤ n ≤ 2 ⋅ 1 0 5 1 \le n \le 2 \cdot 10^5 1≤n≤2⋅105) --- the length of a a a.
The second line contains n n n integers a 1 , a 2 , ... , a n a_1, a_2, \ldots, a_n a1,a2,...,an ( 1 ≤ a i ≤ 1 0 9 1 \le a_i \le 10^9 1≤ai≤109) --- the starting values of array a a a.
It is guaranteed that the sum of n n n over all test cases will not exceed 2 ⋅ 1 0 5 2 \cdot 10^5 2⋅105.
Output
For each test case, output n n n integers representing the answer for each prefix of a a a modulo 1 0 9 + 7 10^9+7 109+7.
Example
| input |
|---|
| 3 |
| 10 |
| 1 2 3 4 5 6 7 8 9 10 |
| 11 |
| 1 6 9 4 7 4 4 10 3 2 3 |
| 4 |
| 527792568 502211460 850237282 374773208 |
| output |
|---|
| 1 3 8 13 46 59 126 149 1174 1311 |
| 1 7 22 26 70 74 150 1303 1306 1308 1568 |
| 527792568 83665723 399119771 773892979 |
note
For each prefix in the first example, a possible array after operations is:
- [ 1 ] [1] [1] and the sum is 1 1 1;
- [ 1 , 2 ] [1, 2] [1,2] and the sum is 3 3 3;
- [ 1 , 1 , 6 ] [1, 1, 6] [1,1,6] and the sum is 8 8 8;
- [ 1 , 1 , 3 , 8 ] [1, 1, 3, 8] [1,1,3,8] and the sum is 13 13 13;
- [ 1 , 1 , 3 , 1 , 40 ] [1, 1, 3, 1, 40] [1,1,3,1,40] and the sum is 46 46 46;
- [ 1 , 1 , 3 , 1 , 5 , 48 ] [1, 1, 3, 1, 5, 48] [1,1,3,1,5,48] and the sum is 59 59 59;
- [ 1 , 1 , 3 , 1 , 5 , 3 , 112 ] [1, 1, 3, 1, 5, 3, 112] [1,1,3,1,5,3,112] and the sum is 126 126 126;
- [ 1 , 1 , 3 , 1 , 5 , 3 , 7 , 128 ] [1, 1, 3, 1, 5, 3, 7, 128] [1,1,3,1,5,3,7,128] and the sum is 149 149 149;
- [ 1 , 1 , 3 , 1 , 5 , 3 , 7 , 1 , 1152 ] [1, 1, 3, 1, 5, 3, 7, 1, 1152] [1,1,3,1,5,3,7,1,1152] and the sum is 1174 1174 1174;
- [ 1 , 1 , 3 , 1 , 5 , 3 , 7 , 1 , 9 , 1280 ] [1, 1, 3, 1, 5, 3, 7, 1, 9, 1280] [1,1,3,1,5,3,7,1,9,1280] and the sum is 1311 1311 1311.
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 mod = 1E9 + 7;
int ksm(int a, int b) {
int res = 1;
while (b) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
void solve() {
int n;
cin >> n;
std::vector<int> a(n), b(n), c(n, 0), d(n, 0);
for (int i = 0; i < n; i ++) {
cin >> a[i], b[i] = a[i];
while (b[i] % 2 == 0) c[i] ++, b[i] /= 2;
d[i] = b[i];
}
for (int i = 1; i < n; i ++) c[i] += c[i - 1], (d[i] += d[i - 1]) %= mod;
int sum = 0;
stack<int> stk;
for (int i = 0; i < n; i ++) {
auto comp = [&](int a, int b, int c) -> int {
// a </> b * 2^c
if (c > 30) return 1;
return a <= b * (1ll << c);
};
while (stk.size() && comp(b[stk.top()], b[i], c[i] - c[stk.top()])) {
int lst = 0, pos = stk.top();
stk.pop();
if (stk.size()) lst = c[stk.top()];
sum -= (b[pos] * ksm(2, c[pos] - lst) % mod - b[pos]);
}
// cout << sum << " ";
if (stk.size()) sum += b[i] * ksm(2, c[i] - c[stk.top()]) % mod - b[i];
else sum += b[i] * ksm(2, c[i]) % mod - b[i];
// cout << sum << " ";
stk.push(i);
cout << ((sum % mod) + d[i] + mod) % mod << " ";
}
cout << endl;
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}E. Monster
Problem Statement
Man, this Genshin boss is so hard. Good thing they have a top-up of 6 6 6 coins for only $ 4.99 . I should be careful and spend no more than I need to, lest my mom catches me...
You are fighting a monster with z z z health using a weapon with d d d damage. Initially, d = 0 d=0 d=0. You can perform the following operations.
- Increase d d d --- the damage of your weapon by 1 1 1, costing x x x coins.
- Attack the monster, dealing d d d damage and costing y y y coins.
You cannot perform the first operation for more than k k k times in a row.
Find the minimum number of coins needed to defeat the monster by dealing at least z z z damage.
Input
The first line contains a single integer t t t ( 1 ≤ t ≤ 100 1\le t\le 100 1≤t≤100) --- the number of test cases.
The only line of each test case contains 4 integers x x x, y y y, z z z, and k k k ( 1 ≤ x , y , z , k ≤ 1 0 8 1\leq x, y, z, k\leq 10^8 1≤x,y,z,k≤108) --- the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.
Output
For each test case, output the minimum number of coins needed to defeat the monster.
Example
| input |
|---|
| 4 |
| 2 3 5 5 |
| 10 20 40 5 |
| 1 60 100 10 |
| 60 1 100 10 |
| output |
|---|
| 12 |
| 190 |
| 280 |
| 160 |
note
In the first test case, x = 2 x = 2 x=2, y = 3 y = 3 y=3, z = 5 z = 5 z=5, and k = 5 k = 5 k=5. Here's a strategy that achieves the lowest possible cost of 12 12 12 coins:
- Increase damage by 1 1 1, costing 2 2 2 coins.
- Increase damage by 1 1 1, costing 2 2 2 coins.
- Increase damage by 1 1 1, costing 2 2 2 coins.
- Attack the monster, dealing 3 3 3 damage, costing 3 3 3 coins.
- Attack the monster, dealing 3 3 3 damage, costing 3 3 3 coins.
You deal a total of 3 + 3 = 6 3 + 3 = 6 3+3=6 damage, defeating the monster who has 5 5 5 health. The total number of coins you use is 2 + 2 + 2 + 3 + 3 = 12 2 + 2 + 2 + 3 + 3 = 12 2+2+2+3+3=12 coins.
In the second test case, x = 10 x = 10 x=10, y = 20 y = 20 y=20, z = 40 z = 40 z=40, and k = 5 k = 5 k=5. Here's a strategy that achieves the lowest possible cost of 190 190 190 coins:
- Increase damage by 5 5 5, costing 5 ⋅ x 5\cdot x 5⋅x = 50 50 50 coins.
- Attack the monster once, dealing 5 5 5 damage, costing 20 20 20 coins.
- Increase damage by 2 2 2, costing 2 ⋅ x 2\cdot x 2⋅x = 20 20 20 coins.
- Attack the monster 5 5 5 times, dealing 5 ⋅ 7 = 35 5\cdot 7 = 35 5⋅7=35 damage, costing 5 ⋅ y 5\cdot y 5⋅y = 100 100 100 coins.
You deal a total of 5 + 35 = 40 5 + 35 = 40 5+35=40 damage, defeating the monster who has exactly 40 40 40 health. The total number of coins you use is 50 + 20 + 20 + 100 = 190 50 + 20 + 20 + 100 = 190 50+20+20+100=190 coins.
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 x, y, z, k;
cin >> x >> y >> z >> k;
int res = 1E18;
for (int a = 0; a <= 20000; a ++) {
int A = z - (a + 1) * a * k / 2;
if (A < 0) break;
int en;
for (int j = max(1ll, a * k); j <= min(A, a * k + k); j = en + 1) {
if (j == A) en = A;
else en = (A - 1) / ((A - 1) / j);
// cout << a << " " << j - a * k << " " << (A + j - 1) / j << ":" << a * (k * x + y) + (j - a * k) * x + (A + j - 1) / j * y << endl;
res = min(res, a * (k * x + y) + (j - a * k) * x + (A + j - 1) / j * y);
}
}
cout << res << endl;
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int dt;
cin >> dt;
while (dt --)
solve();
return 0;
}F. Tree Operations
Problem Statement
This really says a lot about our society.
One day, a turtle gives you a tree with n n n nodes rooted at node x x x. Each node has an initial nonnegative value; the i i i-th node has starting value a i a_i ai.
You want to make the values of all nodes equal to 0 0 0. To do so, you will perform a series of operations on the tree, where each operation will be performed on a certain node. Define an operation on node u u u as choosing a single node in u u u's subtree ∗ ^{\text{∗}} ∗ and incrementing or decrementing its value by 1 1 1. The order in which operations are performed on nodes is as follows:
- For 1 ≤ i ≤ n 1 \le i \le n 1≤i≤n, the i i i-th operation will be performed on node i i i.
- For i > n i > n i>n, the i i i-th operation will be performed on the same node as operation i − n i - n i−n.
More formally, the i i i-th operation will be performed on the ( ( ( i − 1 ) m o d n ) + 1 ) (((i - 1) \bmod n) + 1) (((i−1)modn)+1)-th node. † ^{\text{†}} †
Note that you cannot skip over operations; that is, you cannot perform the i i i-th operation without first performing operations 1 , 2 , ... , i − 1 1, 2, \ldots, i - 1 1,2,...,i−1.
Find the minimum number of operations you must perform before you can make the values of all nodes equal to 0 0 0, assuming you pick operations optimally. If it's impossible to make the values of all nodes equal to 0 0 0 after finite operations, output − 1 -1 −1.
∗ ^{\text{∗}} ∗The subtree of a node u u u is the set of nodes for which u u u lies on the shortest path from this node to the root, including u u u itself.
† ^{\text{†}} †Here, a m o d b a \bmod b amodb denotes the remainder from dividing a a a by b b b.
Input
The first line contains a single integer t t t ( 1 ≤ t ≤ 100 1\le t\le 100 1≤t≤100) --- the number of test cases.
The first line of each test case contains two integers n n n and x x x ( 1 ≤ n ≤ 2000 1 \le n \le 2000 1≤n≤2000, 1 ≤ x ≤ n 1 \le x \le n 1≤x≤n) --- the number of nodes and the root of the tree.
The second line of each test case contains n n n integers a 1 , a 2 , ... , a n a_1, a_2, \ldots, a_n a1,a2,...,an ( 0 ≤ a i ≤ 1 0 9 0 \le a_i \le 10^9 0≤ai≤109) --- the starting value of each node.
Each of the next n − 1 n - 1 n−1 lines of each test case contains two integers u u u and v v v ( 1 ≤ u , v ≤ n 1 \le u, v \le n 1≤u,v≤n, u ≠ v u \neq v u=v) representing an undirected edge from u u u to v v v. It is guaranteed that the given edges form a tree.
It is guaranteed that the sum of n n n over all test cases does not exceed 2000 2000 2000.
Output
For each test case, output a single integer denoting the minimum amount of operations needed to make all nodes 0 0 0. If it's impossible to make all nodes 0 0 0, output − 1 -1 −1.
Example
| input |
|---|
| 5 |
| 2 1 |
| 1 2 |
| 1 2 |
| 3 2 |
| 2 1 3 |
| 2 1 |
| 3 2 |
| 4 1 |
| 1 1 0 1 |
| 1 2 |
| 2 3 |
| 1 4 |
| 12 6 |
| 14 4 5 6 12 9 5 11 6 2 1 12 |
| 3 9 |
| 10 6 |
| 6 12 |
| 4 3 |
| 3 1 |
| 5 11 |
| 9 7 |
| 5 6 |
| 1 8 |
| 2 8 |
| 5 1 |
| 1 1 |
| 0 |
| output |
|---|
| 3 |
| 6 |
| 5 |
| 145 |
| 0 |
Solution
具体见文后视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 2E3 + 10;
int n, rt;
int a[N], ord[N], idx;
LL dp[N];
std::vector<int> g[N];
void DFS(int u, int fa) {
for (auto v : g[u]) {
if (v == fa) continue;
DFS(v, u);
}
ord[ ++ idx] = u;
}
void solve() {
cin >> n >> rt;
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);
}
idx = 0, DFS(rt, -1);
LL res = 1E18;
for (int i = 0; i < 2 * n; i ++ ) {
LL lo = 0, ro = 1E9, cnt = 1E18;
while (lo <= ro) {
LL mid = lo + ro >> 1, x = mid * 2 * n + i;
for (int i = 1; i <= n; i ++) dp[i] = 0;
for (int i = 1; i <= n; i ++) {
int u = ord[i];
for (auto v : g[u]) dp[u] += dp[v];
dp[u] += a[u] - (x / n + (x % n >= u));
if (dp[u] < 0) dp[u] = (-dp[u]) % 2;
}
if (dp[rt] == 0) ro = mid - 1, cnt = x;
else lo = mid + 1;
}
res = min(res, cnt);
}
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;
}视频讲解
Codeforces Global Round 27（A ~ F 题讲解）
最后祝大家早日