24 Game（数学构造）

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.

Sample test(s)

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

 

       题意：

       给出 N（1 ~ 100000），代表有一个 1 ~ N 的数列的数给你用，每次挑出来两个数，进行 a + b 或者 a * b 或者 a - b，将得数加入到数列中，问能否找出一种方法满足最后数列只存在一个数 24，不能则输出 NO，能则输出 YES，且输出运算过程。

 

       思路：

       数学构造。写出来会发现，1 ~ 3 是不可能构成的，根据 4 和 5 两个特殊的序列构成之后，后面的数可以运用 ai - ai-1 = 1 ，1 * 24 = 24，这样的式子来构造出解，最后输出即可。

 

       AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int main() {

    int n;
    scanf("%d", &n);

    if (n <= 3) printf("NO\n");
    else {
        printf("YES\n");

        int m;

        if (n % 2) m = n - 5;
        else m = n - 4;

        m /= 2;

        for (int i = 1; i <= m; ++i) {
            printf("%d - %d = 1\n", n, n - 1);
            n -= 2;
        }

        if (n == 5) {
            printf("5 - 2 = 3\n");
            printf("3 - 1 = 2\n");
            printf("4 * 3 = 12\n");
            printf("12 * 2 = 24\n");
        } else {
            printf("1 * 2 = 2\n");
            printf("2 * 3 = 6\n");
            printf("6 * 4 = 24\n");
        }


        for (int i = 1; i <= m; ++i) {
            printf("1 * 24 = 24\n");
        }

    }

    return 0;
}