• 博客园logo
  • 会员
  • 众包
  • 新闻
  • 博问
  • 闪存
  • 赞助商
  • HarmonyOS
  • Chat2DB
    • 搜索
      所有博客
    • 搜索
      当前博客
  • 写随笔 我的博客 短消息 简洁模式
    用户头像
    我的博客 我的园子 账号设置 会员中心 简洁模式 ... 退出登录
    注册 登录

  • 博客园
  • 联系
  • 订阅
  • 管理

View Post

CodeForces 468A Program F

Description

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: "aopb = 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 Input

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



 

  • using namespace std;  
  • int main()  
  • {  
  • int n;  
  • while(~scanf("%d",&n))  
  •     {  
  • if (n < 4)  
  •         {  
  •             printf("NO");  
  • continue;  
  •         }  
  •         printf("YES\n");  
  • if (n&1)//奇数  
  •         {  
  •             printf("3 + 4 = 7\n");  
  •             printf("7 + 5 = 12\n");  
  •             printf("12 * 2 = 24\n");  
  •             printf("24 * 1 = 24\n");  
  •         }  
  • else  
  •         {  
  •             printf("1 * 2 = 2\n");  
  •             printf("2 * 3 = 6\n");  
  •             printf("6 * 4 = 24\n");  
  •         }  
  • for (int i = n; i > 5; i -= 2)//全部转化为1  
  •         {  
  •             printf("%d - %d = 1\n",i,i-1);  
  •             printf("24 * 1 = 24\n");  
  •         }  
  •     }  
  • return 0;  
  • }  

posted on 2015-07-26 20:10  tony-cao  阅读(140)  评论(0)    收藏  举报

刷新页面返回顶部
 
博客园  ©  2004-2025
浙公网安备 33010602011771号 浙ICP备2021040463号-3