CF 268C 24 Game

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

解析

刚拿到题目觉得完全没有思路。其实后来发现就像脑筋急转弯。

手动前几组

1：NO
2：NO
3：NO
4：YES 24=1*2*3*4

5：YES 24=3*4*(5-2-1)

6：YES 24=(6-5)*(1*2*3*4)

7：YES 24=(7-6)*[3*4*(5-2-1)]

8：YES 24=(8-7)*(6-5)*(1*2*3*4)

9：YES 24=(9-8)*(7-6)*[3*4*(5-2-1)]

……

#include<cstdio>

using namespace std;

int N;

void work()
{
if(N<=3){printf("NO\n"); return;}
printf("YES\n");
if(N%2==0)
{
printf("1 * 2 = 2\n");
printf("3 * 2 = 6\n");
printf("4 * 6 = 24\n");
for(int i=5;i<=N;i+=2)
{
printf("%d - %d = 1\n",i+1,i);
printf("1 * 24 = 24\n");
}
return;
}
if(N%2==1)
{
printf("3 * 4 = 12\n");
printf("5 - 1 = 4\n");
printf("4 - 2 = 2\n");
printf("2 * 12 = 24\n");
for(int i=6;i<=N;i+=2)
{
printf("%d - %d = 1\n",i+1,i);
printf("1 * 24 = 24\n");
}
return;
}
}

int main()
{
//freopen("C.in","r",stdin);

while(scanf("%d",&N)==1)
work();

//while(1);
return 0;
}