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

$frac{n + n + n + n}{n} imes frac{n + n + n + n + n + n}{n} = 24$

 

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 = 2100;
const char str[16][maxn] = {
    "-1",
    "-1",
    "-1",
    "-1",
    "1 * 2
5 + 3
6 + 4",
    "1 * 2
6 * 3
7 - 4
8 / 5",
    "1 + 2
7 + 3
8 + 4
9 + 5
10 - 6",
    "1 + 2
8 + 3
4 + 5
10 + 6
11 / 7
9 + 12",
    "1 + 2
9 + 3
4 + 5
11 - 6
12 - 7
13 / 8
10 + 14",
    "1 + 2
10 + 3
4 + 5
12 + 6
13 / 7
11 - 14
15 - 8
16 + 9",
    "1 + 2
3 + 4
12 + 5
13 + 6
14 / 7
11 + 15
8 - 9
17 / 10
16 + 18",
    "1 + 2
3 + 4
13 / 5
12 + 14
15 - 6
16 + 7
17 - 8
18 + 9
19 - 10
20 + 11",
    "1 + 2
3 + 13
4 + 14
5 + 6
7 + 16
8 + 17
9 + 15
10 + 19
18 / 11
20 / 12
21 * 22",
    "1 + 2
3 + 4
15 / 5
14 - 16
17 - 6
18 + 7
19 - 8
20 + 9
21 - 10
22 + 11
23 - 12
24 + 13",
    "1 + 2
3 + %d
4 + %d
5 + 6
7 + %d
8 + %d
9 + %d
10 + %d
%d / 11
%d / 12
%d * %d
"
};
int main() {
    int n;
    while(~scanf("%d",&n)) {
        if(n <= 13) puts(str[n]);
        else {
            printf(str[14],n+1,n+2,n+4,n+5,n+3,n+7,n+6,n+8,n+9,n+10);
            int last = n + 12;
            printf("%d - %d
",13,14);
            for(int i = 15; i <= n; ++i)
                printf("%d * %d
",i,last++);
            printf("%d + %d
",n + 11,last);
        }
    }
    return 0;
}