Codeforces 469C 24 Game【思维+模拟】

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.

Examples

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

题目大意：

给你N个数字，从1~N.让你通过+ - *的n-1个式子，将所有数都进行运算得到结果是24.

详情移至原题.

思路：

①首先，如果n<4.是一定得不到结果的。

②如果n==4.那么我们答案可以通过：

3 * 4 =12

1 * 2 = 2

12 * 2 = 24得到。

现在考虑如果n==8.那么我们其实就是多了 5 6 7 8这四个数，显然，这四个数是可以相互抵消的：

6 - 5 = 1
7 - 8 = -1
1 + -1 = 0

那么对于最终结果的影响为 ： 24 + 0 = 24.那么显然n%4==0的结果都是可以通过依次类推得到。

③接下来考虑n==5的情况，我们也不难发现，我们也可以通过5以内的所有数字来得到24：

3 * 5 = 15

2 * 4 = 8
15 + 8 =23

23 + 1 =25

那么同理，对于n%4==1的情况，我们也可以以此类推得到。

那么n==6.以及n==7的情况，也是可以同理搞掉的。

确定了思路，接下来的任务就是要去模拟代码了。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int n;
while(~scanf("%d",&n))
{
if(n<4)printf("NO\n");
else
{
if(n%2==1)
{
printf("YES\n");
if(n%4==1)
{
printf("3 * 5 = 15\n");
printf("2 * 4 = 8\n");
printf("15 + 8 =23\n");
printf("23 + 1 = 24\n");
int tmp=(n-5)/2;
int ans=24;
int pre=1;
for(int i=6;i<=n;i+=2)
{
if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
}
for(int i=1;i<=tmp;i++)
{
if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
else printf("%d + -1 = %d\n",ans,ans-1),ans--;
}
}
else
{
printf("4 * 6 = 24\n");
printf("7 - 5 = 2\n");
printf("2 - 1 = 1\n");
printf("3 - 2 = 1\n");
printf("1 - 1 = 0\n");
printf("24 + 0 = 24\n");
int tmp=(n-7)/2;
int ans=24;
int pre=1;
for(int i=8;i<=n;i+=2)
{
if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
}
for(int i=1;i<=tmp;i++)
{
if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
else printf("%d + -1 = %d\n",ans,ans-1),ans--;
}
}
}
else
{
printf("YES\n");
if(n%4==0)
{
int pre=1;
for(int i=1;i<=n;i+=2)
{
if(i==1||i==3)printf("%d * %d = %d\n",i,i+1,i*(i+1));
else if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
}
int tmp=(n-4)/2;
int ans=24;
printf("2 * 12 = 24\n");
for(int i=1;i<=tmp;i++)
{
if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
else printf("%d + -1 = %d\n",ans,ans-1),ans--;
}
}
else
{
printf("4 * 6 = 24\n");
printf("5 - 3 = 2\n");
printf("2 - 1 = 1\n");
int pre=1;
for(int i=7;i<=n;i+=2)
{
if(pre==1)printf("%d - %d = %d\n",i+1,i,i+1-i),pre=-1;
else printf("%d - %d = %d\n",i,i+1,i-i-1),pre=1;
}
printf("2 - 1 = 1\n");
printf("24 * 1 = 24\n");
int tmp=(n-6)/2;
int ans=24;
for(int i=1;i<=tmp;i++)
{
if(i%2==0)printf("%d + 1 = %d\n",ans,ans+1),ans++;
else printf("%d + -1 = %d\n",ans,ans-1),ans--;
}
}
}
}
}
}