Codeforces Round #813 (Div. 2)

A.Wonderful Permutation

题目描述

God's Blessing on This PermutationForces!

A Random Pebble

You are given a permutation p_1,p_2,\ldots,p_n of length n and a positive integer k ≤ n .

In one operation you can choose two indices i and j ( 1 ≤ i < j ≤ n ) and swap p_i with p_j .

Find the minimum number of operations needed to make the sum p_1 + p_2 + \ldots + p_k as small as possible.

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation ( 2 appears twice in the array) and [1,3,4] is also not a permutation ( n=3 but there is 4 in the array).

输入格式

Each test contains multiple test cases. The first line contains the number of test cases t ( 1 ≤ t ≤ 100 ). Description of the test cases follows.

The first line of each test case contains two integers n and k ( 1 ≤ k ≤ n ≤ 100 ).

The second line of each test case contains n integers p_1,p_2,\ldots,p_n ( 1 ≤ p_i ≤ n ). It is guaranteed that the given numbers form a permutation of length n .

输出格式

For each test case print one integer — the minimum number of operations needed to make the sum p_1 + p_2 + \ldots + p_k as small as possible.

样例 #1

样例输入 #1

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

样例输出 #1

1
0
2
0

提示

In the first test case, the value of p_1 + p_2 + \ldots + p_k is initially equal to 2 , but the smallest possible value is 1 . You can achieve it by swapping p_1 with p_3 , resulting in the permutation [1, 3, 2] .

In the second test case, the sum is already as small as possible, so the answer is 0 .

题意

给你一串由1-n组成的数组（长度为n）,和一个≤n的数k。
要求你替换p[1] - p[k]之间的数，使得p[1] 到 p[k]的和最小

思路

我们可以知道在该数组中只有1到n组成，并且没有重复的，因此，要是前k个数的和最小，就要保证前k个数都是小于等于k的。
这样，我们就可以从k+1开始寻找，如果这个数大于k，ans++

C++代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int N = 110;

int n , k;
int p[N];


void solve()
{
	cin >> n >> k;
	for(int i = 1;i <= n;i ++)
		cin >> p[i];
	
	int ans = 0;
	for(int i = k + 1;i <= n;i ++)
	{
		if(p[i] <= k)
			ans ++;
	}
	
	cout << ans << endl;
}

int main()
{
	int T;
	cin >> T;
	
	while(T --) solve();
	
	return 0;
}

B.Woeful Permutation

题目描述

I wonder, does the falling rain

Forever yearn for it's disdain?

Effluvium of the Mind

You are given a positive integer n .

Find any permutation p of length n such that the sum \operatorname{lcm}(1,p_1) + \operatorname{lcm}(2, p_2) + \ldots + \operatorname{lcm}(n, p_n) is as large as possible.

Here \operatorname{lcm}(x, y) denotes the least common multiple (LCM) of integers x and y .

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation ( 2 appears twice in the array) and [1,3,4] is also not a permutation ( n=3 but there is 4 in the array).

输入格式

Each test contains multiple test cases. The first line contains the number of test cases t ( 1 ≤ t ≤ 1\,000 ). Description of the test cases follows.

The only line for each test case contains a single integer n ( 1 ≤ n ≤ 10^5 ).

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

输出格式

For each test case print n integers p_1 , p_2 , \ldots , p_n — the permutation with the maximum possible value of \operatorname{lcm}(1,p_1) + \operatorname{lcm}(2, p_2) + \ldots + \operatorname{lcm}(n, p_n) .

If there are multiple answers, print any of them.

样例 #1

样例输入 #1

2
1
2

样例输出 #1

1 
2 1

提示

For n = 1 , there is only one permutation, so the answer is [1] .

For n = 2 , there are two permutations:

• [1, 2] — the sum is \operatorname{lcm}(1,1) + \operatorname{lcm}(2, 2) = 1 + 2 = 3 .
• [2, 1] — the sum is \operatorname{lcm}(1,2) + \operatorname{lcm}(2, 1) = 2 + 2 = 4 .

题意

给你一个n，要求你创建一个有1到n组成的数组，该数组中，没有元素相等，并且要求该数组的lcm(1,p1) + lcm(2,p2) + ... + lcm(n,pn)最大

思路

我们不难发现，lcm(i,p[i-1]) + lcm(i-1,p[i])最大
因此：

• 如果n为偶数时，刚好可以两两一组交换
• 如果n为奇数时，我们将p[1]匹配1，剩下的元素两两一组

C++代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int N = 1e5+10;

int n;
int p[N];


void solve()
{
	cin >> n ;
	
	for(int i = n;i >= 2;i -= 2)
	{
		p[i] = i - 1;
		p[i - 1] = i;
	}
	
	if(n % 2 == 1)//奇数的话，第一项会漏掉 
		p[1] = 1;
	 
	for(int i = 1;i <= n;i ++)
		cout << p[i] << " ";
	
	cout << endl;
}

int main()
{
	int T;
	cin >> T;
	
	while(T --) solve();
	
	return 0;
}

C.Sort Zero

题目描述

An array is sorted if it has no inversions

A Young Boy

You are given an array of n positive integers a_1,a_2,\ldots,a_n .

In one operation you do the following:

1. Choose any integer x .
2. For all i such that a_i = x , do a_i := 0 (assign 0 to a_i ).

Find the minimum number of operations required to sort the array in non-decreasing order.

输入格式

Each test contains multiple test cases. The first line contains the number of test cases t ( 1 ≤ t ≤ 10^4 ). Description of the test cases follows.

The first line of each test case contains a single integer n ( 1 ≤ n ≤ 10^5 ).

The second line of each test case contains n positive integers a_1,a_2,\ldots,a_n ( 1 ≤ a_i ≤ n ).

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

输出格式

For each test case print one integer — the minimum number of operations required to sort the array in non-decreasing order.

样例 #1

样例输入 #1

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

样例输出 #1

1
2
4
3
0

提示

In the first test case, you can choose x = 3 for the operation, the resulting array is [0, 0, 2] .

In the second test case, you can choose x = 1 for the first operation and x = 3 for the second operation, the resulting array is [0, 0, 0, 0] .

题意

给你一个长度为n的数组a，每一次可以进行一个操作：选择一个数x,将数组a中所有等于x的数变为0，求将这个数组变成不下降序列，最少需要操作多少次？

思路

我们用set容器维护1~i值的个数，如果出现a[i] > a[i+1]，那么就将1~i中的所有值标为0，答案更新为1 ~ i中值的个数

C++代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <set>

using namespace std;

const int N = 1e5+10;

int n;
int a[N] ;
bool st[N];

void solve()
{
	cin >> n;
	
	for(int i = 1;i <= n;i ++)
		cin >> a[i];
	
	memset(st,false,sizeof st);
	
	set<int> s;
	int ans = 0,last = 0;
	
	for(int i = 1;i < n;i ++)
	{
		if(st[ a[i] ] == true)
			a[i] = 0;
		if(st[ a[i + 1] ] == true)
			a[i + 1] = 0;
			
		if(a[ i ] != 0)
			s.insert(a[i]);
			
		if(a[ i ] > a[i + 1])
		{
			for(int j = last;j <= i;j ++)
				st[ a[j] ] = true;
				
			last = i;
			
			ans = s.size();	
		}
	}
	
//	cout << "Ans = ";
	cout << ans << endl;
	
}

int main()
{
	int T;
	cin >> T;
	
	while(T --) solve();
	
	return 0;
}