Codeforces Round #756 (Div. 3)

题目如下：

Polycarp has an integer

Reverse the prefix of length

Note that for different operations, the values of

Polycarp loves even numbers. Therefore, he wants to make his number even. At the same time, Polycarp is very impatient. He wants to do as few operations as possible.

Help Polycarp. Determine the minimum number of operations he needs to perform with the number

You need to answer

Input

The first line contains the number

Each of the following

Output

Note

In the first test case,

In the second test case,

Select
Select

It can be shown that

In the third test case,

In the fourth test case,

题意很简单，一个数，每次都让其前n个数字倒序，问最少几次能让它变成一个偶数。

很显然，有四种情况，一、这个数本身是偶数，输出0.二、这个数每一位数字都是奇数，那么不可能能转成偶数，输出-1。三、这个数字的第一位数字为偶数，最后一位为奇数，那么直接整个倒序即可，输出1.四、头尾均为奇数，中间有偶数，那么先通过一次倒序把这个偶数弄到第一位，然后再一次转到最后，也就是2次，输出2。总而言之水题，代码：

#include <iostream>
using namespace std;
 
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        string str;
        cin >> str;
        int temp = -1;
        int n;
        for (int i = 0; i < str.size(); i++)
        {
            n = str[i] - '0';
            if (n % 2 == 0)
            {
                temp = 2;
                break;
            }
        }
        n = str[0] - '0';
        if (n % 2 == 0)
        {
            temp = 1;
        }
        n = str[str.size() - 1] - '0';
        if (n % 2 == 0)
        {
            temp = 0;
        }
        cout << temp << endl;
    }
    return 0;
}

The All-Berland Team Programming Contest will take place very soon. This year, teams of four are allowed to participate.

There are

each team must consist of exactly
teams of

Thus, each team must have at least one programmer and at least one mathematician.

Print the required maximum number of teams. Each person can be a member of no more than one team.

Input

The first line contains an integer

This is followed by descriptions of

Output

Note

In the first test case of the example, two teams can be composed. One way to compose two teams is to compose two teams of

In the second test case of the example, only one team can be composed:

输入两个数代表科学家和数学家的数量，他们要四个人一组去旅游，这四个人中间必须要有一个数学家一个科学家才行，问一共能组成几个队。

也是水题，比较2个值就行了，一个是(a+b)/4，还有一个是两个中的较小值，输出这两个数的较小值就是答案。

代码：

#include <iostream>
 #include <cstring> 
 #include <cstdio>
 #include <algorithm>
 
 using namespace std;
 
 int main()
{
    int t;
    cin>>t;
    int a,b;
    int temp;
    while(t--)
        {
        cin>>a>>b;
        temp=(a+b)/4;
        if(a>b)swap(a,b);
        if(temp>=a)cout<<a;
        else cout<<temp;
        cout<<endl;
    }
 }

Polycarp wrote on a whiteboard an array

He also prepared a resulting array

After that, he did exactly

Look at the leftmost and rightmost elements of
If you picked the leftmost element of
The picked element is erased from

Note that on the last step,

Let's look at an example. Let

During the first step, the minimum is on the right (with a value of
During the second step, the minimum is on the left (with a value of
During the third step, the minimum is on the left (with a value of
During the fourth step, the minimum is both left and right (this value is

Thus, a possible value of

You are given the final value of the resulting array

Input

The first line of the input contains an integer

Each test case consists of two lines. The first of them contains an integer

It is guaranteed that the sum of the values

Output

题意分析及解题思路：

这是道很有意思的题目，首先要了解以下的概念，给你一个由1到n组成的乱序数字，如n=4，p=[3 1 4 2],接下来将p数组中的头和尾元素进行比较，2<3，并且2在p的右边，那么就把2这个元素插入一个新数组a=[](初始为空)的右边，a=[2]，接下来p=[3 1 4],3<4,并且3在p的左边，把3插入a的左边，a=[3 2]。如此以往，直到p中只剩下一个元素4，把它放在a的左边或者右边都可以，也就是a=[1 3 2 4]或a=[4 1 3 2]都是可以的。

了解了这个概念以后，现在给你转换后的a数组，问你能不能找出初始的p数组，经过这个转换以后能变成现在的a数组。

首先要排除暴力反推的方法，因为肯定会超时(经常尝试也确实是超时了)，所以这题肯定是通过找规律的方法做的

确定这个思路以后，我们首先要思考，什么样的数组a是没有对应的p的？

认真观察推导a的过程，我们不难看出啊，每次都是找出小的那个值插到a里面去的，这说明什么？说明什么？说明最大的值肯定会留到最后，也就是最大的值肯定在a数组的头或者尾。那么只要头和尾都不是n的a，肯定是没有对应的p的。其次，再仔细观察，每次比较之后都是较小值插入进去，较大值留下，其实每个数都是在和大值的比较之后被插入进入，也就是相当于对于最大值逆序了一遍。所以我们只要分最大值在左和在右的两种情况，将其他的a倒序输出，得到的就是对应的p(这个规律比较难理解，需要慢慢通过例子推才能知道)。

代码：

#include <iostream>
using namespace std;
 
int num1[200000];
 
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int b; cin >> b;
 
        for (int i = 0; i < b; i++)
        {
            cin >> num1[i];
        }
        if (num1[b - 1] != b && num1[0] != b)
        {
            cout << -1 << endl; continue;
        }
        if (b == 1 && num1[0] == b) cout << b;
        else if (num1[0] == b) {
            cout << num1[0] << " ";
            for (int i = 0; i < b - 1; i++)
            {
                cout << num1[b - 1 - i] << " ";
            }
            cout << endl;
        }
        else if (num1[b - 1] == b) {
            for (int i = 0; i < b - 1; i++)
            {
                cout << num1[b - 2 - i] << " ";
            }
            cout << num1[b - 1];
        }
        cout << endl;
    }
    return 0;
}

总结：

这场基本都是找规律的题目，分清几种情况，弄出来就行了，就是有些题目的规模比较难想明白，需要慢慢琢磨。

posted @ 2021-11-30 00:15 Sheep灬阅读(142) 评论(0) 编辑收藏举报

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Codeforces Round #756 (Div. 3)

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