Codeforces Round #768 (Div. 2)

B. Fun with Even Subarrays

You are given an array $a$ of $n$ elements. You can apply the following operation to it any number of times:

Select some subarray from a of even size $2k$ that begins at position $l$ $(1≤l≤l+2⋅k−1≤n, k≥1)$ and for each $i$ between $0$ and $k−1$ (inclusive), assign the value $a_l+k+i$ to $a_l+i$.
For example, if $a=[2,1,3,4,5,3]$, then choose $l=1$ and $k=2$, applying this operation the array will become $a=[3,4,3,4,5,3]$.

Find the minimum number of operations (possibly zero) needed to make all the elements of the array equal.

Input

The input consists of multiple test cases. The first line contains a single integer $t (1≤t≤2⋅10^4$) — the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer $n (1≤n≤2⋅10^5)$ — the length of the array.

The second line of each test case consists of $n$ integers $a_1,a_2,…,a_n (1≤a_i≤n)$ — the elements of the array $a$.

It is guaranteed that the sum of n over all test cases does not exceed $2⋅10^5$.

Output

Print $t$ lines, each line containing the answer to the corresponding test case — the minimum number of operations needed to make equal all the elements of the array with the given operation.

Example

input

5
3
1 1 1
2
2 1
5
4 4 4 2 4
4
4 2 1 3
1
1

output

0
1
1
2
0

Note

In the first test, all elements are equal, therefore no operations are needed.

In the second test, you can apply one operation with $k=1$ and $l=1$, set $a_1:=a_2$, and the array becomes $[1,1]$ with $1$ operation.

In the third test, you can apply one operation with $k=1$ and $l=4$, set $a_4:=a_5$, and the array becomes $[4,4,4,4,4]$.

In the fourth test, you can apply one operation with $k=1$ and $l=3$, set $a_3:=a_4$, and the array becomes $[4,2,3,3]$, then you can apply another operation with $k=2$ and $l=1$, set $a_1:=a_3, a_2:=a_4$, and the array becomes $[3,3,3,3]$.

In the fifth test, there is only one element, therefore no operations are needed.

解题思路

思维

每次都是从后面的数字往前面复制相同长度的一段，可知最后数组中相等的元素一定是最后一个数字，可先找出最后数字相等的那段，然后像滚雪球一样往前滚，注意其中有些可能是等于最后一个数字的，这些可以跳过

• 时间复杂度：$O(n)$

代码

// Problem: B. Fun with Even Subarrays
// Contest: Codeforces - Codeforces Round #768 (Div. 2)
// URL: https://codeforces.ml/contest/1631/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms

// %%%Skyqwq
#include <bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }

template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

int t,n,a[200005];
int main()
{
	for(scanf("%d",&t);t;t--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%d",&a[i]);
		int res=0;
		int x=1;
		int pos=n-1;
		while(pos>=1)
		{
			if(pos>=1&&a[pos]!=a[n])
			{
				res++;
				pos=2*pos-n;
			}
			else if(pos>=1&&a[pos]==a[n])pos--;
			
		}
		
		printf("%d\n",res);
	}	
	return 0;
}

C. And Matching

You are given a set of $n$ ($n$ is always a power of $2$) elements containing all integers $0,1,2,…,n−1$ exactly once.

Find $n/2$ pairs of elements such that:

Each element in the set is in exactly one pair.
The sum over all pairs of the bitwise AND of its elements must be exactly equal to $k$. Formally, if $a_i$ and $b_i$ are the elements of the $i-th$ pair, then the following must hold:
$∑_{i=1}^{n/2}a_i\& b_i=k$,
where $\&$ denotes the bitwise AND operation.
If there are many solutions, print any of them, if there is no solution, print $−1$ instead.

Input

The input consists of multiple test cases. The first line contains a single integer $t (1≤t≤400)$ — the number of test cases. Description of the test cases follows.

Each test case consists of a single line with two integers $n$ and $k$ ($4≤n≤2^{16}, n$ is a power of $2$, $0≤k≤n−1$).

The sum of $n$ over all test cases does not exceed $2^{16}. All test cases in each individual input will be pairwise different.

Output

For each test case, if there is no solution, print a single line with the integer $−1$.

Otherwise, print $n/2$ lines, the $i-th4 of them must contain $a_i$ and $b_i$, the elements in the $i-th$ pair.

If there are many solutions, print any of them. Print the pairs and the elements in the pairs in any order.

Example

input

4
4 0
4 1
4 2
4 3

output

0 3
1 2
0 2
1 3
0 1
2 3
-1

Note

In the first test, $(0\&3)+(1\&2)=0.$

In the second test, $(0\&2)+(1\&3)=1.$

In the third test, $(0\&1)+(2\&3)=2.$

In the fourth test, there is no solution.

解题思路

构造

首先需要注意 $k$ 的取值范围，任意两个数的$\&$运算之和不止 $n-1$，对于 $k$ 不妨考虑分类讨论：

• $k=0$ 时，让首尾两个数依次组成一对

• $k>0\, and\, k<n-1$ 时，在上面的基础上，让 $n-1$ 跟 $k$ 组成一对，$0$ 跟 $n-1-k$ 组成一对

• $k=n-1$ 时，

• • $n=4$ 时无解
• • 否则，让 $n-1$ 跟 $n-2$，$1$ 跟 $n-3$，$0$ 跟 $3$组成一对

• 时间复杂度：$O(n)$

代码

// Problem: C. And Matching
// Contest: Codeforces - Codeforces Round #768 (Div. 2)
// URL: https://codeforces.com/contest/1631/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms

// %%%Skyqwq
#include <bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }

template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

int t,n,k;
int main()
{
	for(scanf("%d",&t);t;t--)
	{
		scanf("%d%d",&n,&k);
		if(k==0)
			for(int i=0;i<n/2;i++)printf("%d %d\n",i,n-1-i);
		else if(k>0&&k<n-1)
		{
			printf("%d %d\n",k,n-1);
			printf("0 %d\n",n-1-k);
			for(int i=1;i<n/2;i++)
				if(i!=k&&i!=n-1-k)printf("%d %d\n",i,n-1-i);
		}
		else if(k==n-1)
		{
			if(n<=4)puts("-1");
			else
			{
				puts("0 2");
				printf("%d %d\n",1,n-3);
				printf("%d %d\n",n-2,n-1);
				for(int i=3;i<n/2;i++)printf("%d %d\n",i,n-1-i);
			}
		}
		else
			puts("-1");
	}
	return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/15851909.html
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