2015 Multi-University Training Contest 2 hdu 5308 I Wanna Become A 24-Point Master

I Wanna Become A 24-Point Master

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 897    Accepted Submission(s): 379
Special Judge

Problem Description

Recently Rikka falls in love with an old but interesting game -- 24 points. She wants to become a master of this game, so she asks Yuta to give her some problems to practice.

Quickly, Rikka solved almost all of the problems but the remained one is really difficult:

In this problem, you need to write a program which can get 24 points with n numbers, which are all equal to n.

 

Input

There are no more then 100 testcases and there are no more then 5 testcases with n≥100. Each testcase contains only one integer n (1≤n≤105)

 

Output

For each testcase:

If there is not any way to get 24 points, print a single line with -1.

Otherwise, let A be an array with 2n−1 numbers and at firsrt Ai=n (1≤i≤n). You need to print n−1 lines and the ith line contains one integer a, one char band then one integer c, where 1≤a,c<n+i and b is "+","-","*" or "/". This line means that you let Aa and Ac do the operation b and store the answer into An+i.

If your answer satisfies the following rule, we think your answer is right:

1. A2n−1=24

2. Each position of the array A is used at most one tine.

3. The absolute value of the numerator and denominator of each element in array A is no more than 109

 

Sample Input

4

 

Sample Output

1 * 2

5 + 3

6 + 4

 

Source

2015 Multi-University Training Contest 2

 

 

解题：打表+规律

可以发现当n等于12时，可以求解由
至于其余的数字，假设我们取n = 14 由于得到24，前面n个我们只用到了12个，那么我们可以将13 - 14，然后再加上 24 仍然是24

如果是15 ，我们可以13 - 14，然后差乘以 15 最后积加上24.。。以此类推

$\frac{n + n + n + n}{n} \times \frac{n + n + n + n + n + n}{n} = 24$

 

妈拉个巴子，latex公式不管用了

 

好吧 

 

(n+n+n+n)/n = 4;

(n+n+n+n+n+n)/n = 6;

4*6 = 24

 

所以当n大于12的时候，已经很明显可以解决了

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 const char str[][maxn] = {
     "-1",
     "-1",
     "-1",
     "-1",
     "1 * 2\n5 + 3\n6 + 4",
     "1 * 2\n6 * 3\n7 - 4\n8 / 5",
     "1 + 2\n7 + 3\n8 + 4\n9 + 5\n10 - 6",
     "1 + 2\n8 + 3\n4 + 5\n10 + 6\n11 / 7\n9 + 12",
     "1 + 2\n9 + 3\n4 + 5\n11 - 6\n12 - 7\n13 / 8\n10 + 14",
     "1 + 2\n10 + 3\n4 + 5\n12 + 6\n13 / 7\n11 - 14\n15 - 8\n16 + 9",
     "1 + 2\n3 + 4\n12 + 5\n13 + 6\n14 / 7\n11 + 15\n8 - 9\n17 / 10\n16 + 18",
     "1 + 2\n3 + 4\n13 / 5\n12 + 14\n15 - 6\n16 + 7\n17 - 8\n18 + 9\n19 - 10\n20 + 11",
     "1 + 2\n3 + 13\n4 + 14\n5 + 6\n7 + 16\n8 + 17\n9 + 15\n10 + 19\n18 / 11\n20 / 12\n21 * 22",
     "1 + 2\n3 + 4\n15 / 5\n14 - 16\n17 - 6\n18 + 7\n19 - 8\n20 + 9\n21 - 10\n22 + 11\n23 - 12\n24 + 13",
     "1 + 2\n3 + %d\n4 + %d\n5 + 6\n7 + %d\n8 + %d\n9 + %d\n10 + %d\n%d / 11\n%d / 12\n%d * %d\n"
 };
 int main() {
     int n;
     while(~scanf("%d",&n)) {
         if(n <= ) puts(str[n]);
         else {
             printf(str[],n+,n+,n+,n+,n+,n+,n+,n+,n+,n+);
             int last = n + ;
             printf("%d - %d\n",,);
             for(int i = ; i <= n; ++i)
                 printf("%d * %d\n",i,last++);
             printf("%d + %d\n",n + ,last);
         }
     }
     return ;
 }