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Codeforces Round #268 (Div. 1)A(构造)
2015-07-20 17:37:50 来源: 作者: 【 】 浏览:2
Tags:Codeforces Round #268 Div. 构造
A. 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:用1-n的数字经过n-1次运算得到24
思路:1. n<=3 一定不可以
2. n>=4一定可以,为什么呢
首先注意n=4时,可以通过1*2*3*4=24构造
然后看剩下还有多少个数,如果为偶数个,则前后两两相减得1,
然后就让这些1只做乘运算,最后24 * 1 = 24
如果为奇数怎能办呢
不难发现,n=5时,可以通过4*5+3+2-1=24构造
那么剩下的数的数量一定为偶数了,解法同上
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