24 Game CodeForces - 468A

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.

Example

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

没有思路，以为是搜索，搜题解发现都是固定的“套路”，感觉还是自己的思维不够吧。
＜4肯定组不成24，

>=4,偶数的话最少4个数就可以出来24，奇数的话，5个数就可以，让剩下的数，后一个减去前一个，只保留一个1，然后不断地*24就好了

#include <cstdio>
using namespace std;
int main()
{
int n;
while(~scanf("%d",&n)){
if(n<4){
printf("NO\n");
}
else{
printf("YES\n");
if(n&1){
printf("3 + 4 = 7\n");
printf("7 + 5 = 12\n");
printf("12 * 2 = 24\n");
printf("24 * 1 = 24\n");
while(n>5){
printf("%d - %d =1\n",n,n-1);
printf("24 * 1 =24\n");
n-=2;
}
}
else{
printf("2 * 3 = 6\n");
printf("4 * 6 = 24\n");
printf("24 * 1 = 24\n");
while(n>4){
printf("%d - %d = 1\n",n,n-1);
printf("24 * 1 = 24\n");
n-=2;
}
}
}
}
}