CF468A 24 Game

最新推荐文章于 2024-06-27 20:29:17 发布

ShadyPi

最新推荐文章于 2024-06-27 20:29:17 发布

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本文介绍了一种解决Codeforces上24Game问题的有效方法。针对不同数量的初始整数，给出了具体的策略来最终获得24这个数值。通过简单的数学运算，确保了解的可行性。

原题链接：http://codeforces.com/problemset/problem/468/A

24 Game

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

output

input

output

YES
8 * 7 = 56
6 * 5 = 30
3 - 4 = -1
1 - 2 = -1
30 - -1 = 31
56 - 31 = 25
25 + -1 = 24

题解

一道简单题。

$1,2,3$ 肯定不行，考虑到后面相邻的数两两相减就是 $1$ ，乘的话对答案没有影响，所以们只需要手动把前几个数凑成 $24$ ，后面一直乘 $1$ 就行了。

于是我们可以这么凑：

当 $n$ 为偶数时：

1 \times 2 \times 3 \times 4 = 24

$1\times 2\times 3\times 4=24$

当 $n$ 为奇数时：

1 + 2 \times 4 + 3 \times 5 = 24

$1+2\times 4+3\times 5=24$

大功告成。

代码

#include<bits/stdc++.h>
using namespace std;
int n;
void in(){scanf("%d",&n);}
void ac()
{
    if(n<4){puts("NO");return;}
    puts("YES");
    if(n&1)
    {
        printf("2 * 4 = 8\n3 * 5 = 15\n8 + 15 = 23\n23 + 1 = 24\n");
        for(int i=6;i<n;i+=2)printf("%d - %d = 1\n24 * 1 = 24\n",i+1,i);
        return;
    }
    printf("1 * 2 = 2\n2 * 3 = 6\n6 * 4 = 24\n");
    for(int i=5;i<n;i+=2)printf("%d - %d = 1\n24 * 1 = 24\n",i+1,i);
}
int main(){in();ac();}