Codeforces 469C 24 Game【思维+模拟】

本文介绍了一种基于数字序列操作的游戏——24Game,并详细解析了如何通过加减乘除运算使序列最终结果为24的算法实现。针对不同序列长度n提供了具体的策略和示例代码。

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C. 24 Game
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.

Initially you have a sequence of n integers: 1, 2, ..., n. In a single step, you can pick two of them, let's denote them a and b, erase them from the sequence, and append to the sequence either a + b, or a - b, or a × b.

After n - 1 steps there is only one number left. Can you make this number equal to 24?

Input

The first line contains a single integer n (1 ≤ n ≤ 105).

Output

If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).

If there is a way to obtain 24 as the result number, in the following n - 1 lines print the required operations an operation per line. Each operation should be in form: "a op b = c". Where a and b are the numbers you've picked at this operation; op is either "+", or "-", or "*"; c is the result of corresponding operation. Note, that the absolute value of c mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.

If there are multiple valid answers, you may print any of them.

Examples
Input
1
Output
NO
Input
8
Output
YES
8 * 7 = 56
6 * 5 = 30
3 - 4 = -1
1 - 2 = -1
30 - -1 = 31
56 - 31 = 25
25 + -1 = 24

题目大意:

给你N个数字,从1~N.让你通过+ - *的n-1个式子,将所有数都进行运算得到结果是24.

详情移至原题.


思路:


①首先,如果n<4.是一定得不到结果的。

②如果n==4.那么我们答案可以通过:

3 * 4 =12

1 * 2 = 2

12 * 2 = 24得到。

现在考虑如果n==8.那么我们其实就是多了 5 6 7 8这四个数,显然,这四个数是可以相互抵消的:

6 - 5 = 1

7 - 8 = -1

1 + -1 = 0

那么对于最终结果的影响为 : 24 + 0 = 24.那么显然n%4==0的结果都是可以通过依次类推得到。

③接下来考虑n==5的情况,我们也不难发现,我们也可以通过5以内的所有数字来得到24:

3 * 5 = 15

2 * 4 = 8

15 + 8 =23

23 + 1 =25

那么同理,对于n%4==1的情况,我们也可以以此类推得到。

那么n==6.以及n==7的情况,也是可以同理搞掉的。

确定了思路,接下来的任务就是要去模拟代码了。



Ac代码:


#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n<4)printf("NO\n");
        else
        {
            if(n%2==1)
            {
                printf("YES\n");
                if(n%4==1)
                {
                    printf("3 * 5 = 15\n");
                    printf("2 * 4 = 8\n");
                    printf("15 + 8 =23\n");
                    printf("23 + 1 = 24\n");
                    int tmp=(n-5)/2;
                    int ans=24;
                    int pre=1;
                    for(int i=6;i<=n;i+=2)
                    {
                        if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
                        else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
                    }
                    for(int i=1;i<=tmp;i++)
                    {
                        if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
                        else printf("%d + -1 = %d\n",ans,ans-1),ans--;
                    }
                }
                else
                {
                    printf("4 * 6 = 24\n");
                    printf("7 - 5 = 2\n");
                    printf("2 - 1 = 1\n");
                    printf("3 - 2 = 1\n");
                    printf("1 - 1 = 0\n");
                    printf("24 + 0 = 24\n");
                    int tmp=(n-7)/2;
                    int ans=24;
                    int pre=1;
                    for(int i=8;i<=n;i+=2)
                    {
                        if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
                        else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
                    }
                    for(int i=1;i<=tmp;i++)
                    {
                        if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
                        else printf("%d + -1 = %d\n",ans,ans-1),ans--;
                    }
                }
            }
            else
            {
                printf("YES\n");
                if(n%4==0)
                {
                    int pre=1;
                    for(int i=1;i<=n;i+=2)
                    {
                        if(i==1||i==3)printf("%d * %d = %d\n",i,i+1,i*(i+1));
                        else if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
                        else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
                    }
                    int tmp=(n-4)/2;
                    int ans=24;
                    printf("2 * 12 = 24\n");
                    for(int i=1;i<=tmp;i++)
                    {
                        if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
                        else printf("%d + -1 = %d\n",ans,ans-1),ans--;
                    }
                }
                else
                {
                    printf("4 * 6 = 24\n");
                    printf("5 - 3 = 2\n");
                    printf("2 - 1 = 1\n");
                    int pre=1;
                    for(int i=7;i<=n;i+=2)
                    {
                        if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
                        else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
                    }
                    printf("2 - 1 = 1\n");
                    printf("24 * 1 = 24\n");
                    int tmp=(n-6)/2;
                    int ans=24;
                    for(int i=1;i<=tmp;i++)
                    {
                        if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
                        else printf("%d + -1 = %d\n",ans,ans-1),ans--;
                    }
                }
            }
        }
    }
}






当前提供的引用内容并未提及关于Codeforces比赛M1的具体时间安排[^1]。然而,通常情况下,Codeforces的比赛时间会在其官方网站上提前公布,并提供基于不同时区的转换工具以便参赛者了解具体开赛时刻。 对于Codeforces上的赛事而言,如果一场名为M1的比赛被计划举行,则它的原始时间一般按照UTC(协调世界时)设定。为了得知该场比赛在UTC+8时区的确切开始时间,可以遵循以下逻辑: - 前往Codeforces官网并定位至对应比赛页面。 - 查看比赛所标注的标准UTC起始时间。 - 将此标准时间加上8小时来获取对应的北京时间(即UTC+8)。 由于目前缺乏具体的官方公告链接或者确切日期作为依据,无法直接给出Codeforces M1比赛于UTC+8下的实际发生时段。建议定期访问Codeforces平台查看最新动态更新以及确认最终版程表信息。 ```python from datetime import timedelta, datetime def convert_utc_to_bj(utc_time_str): utc_format = "%Y-%m-%dT%H:%M:%SZ" bj_offset = timedelta(hours=8) try: # 解析UTC时间为datetime对象 utc_datetime = datetime.strptime(utc_time_str, utc_format) # 转换为北京时区时间 beijing_time = utc_datetime + bj_offset return beijing_time.strftime("%Y-%m-%d %H:%M:%S") except ValueError as e: return f"错误:{e}" # 示例输入假设某场Codeforces比赛定于特定UTC时间 example_utc_start = "2024-12-05T17:35:00Z" converted_time = convert_utc_to_bj(example_utc_start) print(f"Codeforces比赛在北京时间下将是:{converted_time}") ```
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