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 than1018.
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 = -11 - 2 = -130 - -1 = 3156 - 31 = 25
25 + -1 = 24

解题说明：此题就是给出一列数算24，可以用暴力求解，构造一个最小集，然后其他的数字相减消去。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
using namespace std;

int main()
{
int n, i;
scanf("%d", &n);
if(n <= 3)
{
printf("NO\n");
return 0;
}
else
{
printf("YES\n");
if(n % 2 == 0)
{
printf("1 * 2 = 2\n");
printf("2 * 3 = 6\n");
printf("6 * 4 = 24\n");
for(i = n; i >= 6; i -= 2)
{
printf("%d - %d = 1\n", i, i-1);
printf("24 * 1 = 24\n");
}
}
else
{
printf("5 - 2 = 3\n");
printf("3 - 1 = 2\n");
printf("2 * 3 = 6\n");
printf("4 * 6 = 24\n");
for(i = n; i >= 7; i -= 2)
{
printf("%d - %d = 1\n", i, i-1);
printf("24 * 1 = 24\n");
}
}
}
return 0;
}