B. GCD Partition
While at Kira's house, Josuke saw a piece of paper on the table with a task written on it.
The task sounded as follows. There is an array $a$ of length $n$. On this array, do the following:
- select an integer $k > 1$;
- split the array into $k$ subsegments $^\dagger$;
- calculate the sum in each of $k$ subsegments and write these sums to another array $b$ (where the sum of the subsegment $(l, r)$ is ${\sum_{j = l}^{r}a_j}$);
- the final score of such a split will be $\gcd(b_1, b_2, \ldots, b_k)^\ddagger$.
The task is to find such a partition that the score is maximum possible. Josuke is interested in this task but is not strong in computer science. Help him to find the maximum possible score.
$^\dagger$ A division of an array into $k$ subsegments is $k$ pairs of numbers $(l_1, r_1), (l_2, r_2), \ldots, (l_k, r_k)$ such that $l_i \le r_i$ and for every $1 \le j \le k - 1$ $l_{j + 1} = r_j + 1$, also $l_1 = 1$ and $r_k = n$. These pairs represent the subsegments.
$^\ddagger$ $\gcd(b_1, b_2, \ldots, b_k)$ stands for the greatest common divisor (GCD) of the array $b$.
Input
The first line contains a single number $t$ ($1 \le t \le 10^4$) — the number of test cases.
For each test case, the first line contains one integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the length of the array $a$.
The second line contains $n$ integers $a_1, a_2, a_3, \ldots, a_n$ ($1 \le a_i \le 10^9 $) — the array $a$ itself.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
Output
For each test case print a single integer — the maximum score for the optimal partition.
Example
input
6 4 2 2 1 3 2 1 2 3 1 4 5 6 1 2 1 1 1 3 10 12 30 37 88 12 78 89 17 2 12 6 7 7 7 7 7 7
output
4 1 5 3 1 21
Note
In the first test case, you can choose $k = 2$ and split the array into subsegments $(1, 2)$ and $(3, 4)$.
Then the score of such a partition will be equal to $\gcd(a_1 + a_2, a_3 + a_4) = \gcd(2 + 2, 1 + 3) = \gcd(4, 4) = 4$.
In the fourth test case, you can choose $k = 3$ and split the array into subsegments $(1, 2), (3, 5), (6, 6)$.
The split score is $\gcd(1 + 2, 1 + 1 + 1, 3) = 3$.
解题思路
可以证明最优解必定出现在将序列分成两段的情况中。
假如将序列分成$m > 2$段,那么此时的答案是$\gcd(b_1, b_2, \ldots, b_m)$,有$\gcd(b_1, b_2, \ldots, b_m) \le \gcd(b_1 + b_2, b_3, \ldots, b_m)$,这是因为对于$b_1, \ldots, b_m$的任意一个公约数$d$,都一定能整除$b_1 + b_2, b_3, \ldots, b_m$,即也是$b_1 + b_2, b_3, \ldots, b_m$的公约数。因此就有$\gcd(b_1, b_2, \ldots, b_m) \le \gcd(b_1 + b_2, b_3, \ldots, b_m)$。以此类推,为了得到最大值最终一定会变成$\gcd(b_1, b_2)$的形式。因此可以枚举两段的分界点来求最大值。
AC代码如下:
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long long LL; 5 6 const int N = 2e5 + 10; 7 8 LL s[N]; 9 10 LL gcd(LL a, LL b) { 11 return b ? gcd(b, a % b) : a; 12 } 13 14 void solve() { 15 int n; 16 scanf("%d", &n); 17 for (int i = 1; i <= n; i++) { 18 // 因为写成scanf("%d", s + i)被hack了,由于有多组测试数据,因此要保证s[i]的64位均为0 19 scanf("%lld", s + i); 20 s[i] += s[i - 1]; 21 } 22 LL ret = 1; 23 for (int i = 1; i < n; i++) { 24 ret = max(ret, gcd(s[i], s[n] - s[i])); 25 } 26 printf("%lld\n", ret); 27 } 28 29 int main() { 30 int t; 31 scanf("%d", &t); 32 while (t--) { 33 solve(); 34 } 35 36 return 0; 37 }
参考资料
Codeforces Round #846 (Div. 2) — Tutorial:https://codeforces.com/blog/entry/111841
标签:10,le,GCD,Partition,score,array,ldots,gcd From: https://www.cnblogs.com/onlyblues/p/17069077.html