CodeForces 469C - 24 Game

Little X used to play a card game called "24 Game", butrecently 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 operationper 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.
Theresult 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

思路：

[1,3]时不能构成24点

4时1*2*3*4=24

5时1*(5-2+3)*4=24

6时2*3*4=24,然后6-5-1=0,再用0去乘大于6的所有数，最后24+0=24

代码：
#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
int n;

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

return 0;
}