HDU 5308 I Wanna Become A 24-Point Master

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=5308

题面：

I Wanna Become A 24-Point Master

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 602    Accepted Submission(s): 249
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 b and 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：
    如此之大的数据量，搜索是肯定不行。但还是被题目那句大于100的数据不会超过5组给蒙了一下。队友之前想着能不能从24往前搜，实则也是不行的。

由于根本不知道前面到底有什么数。又该相应如何的操作。看了题解后，恍然大悟。就应该去构造。

    枚举n比較小的情况，然后当n大于等于14时，能够去凑((4*n)/n)*((6*n)/n),尽管是12个n，可是仍要从14，開始，由于多余的n须要通过一次减法，多次乘法消去。最后再加上之前算出的24就可以。


代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
	int n,tmp;
	while(~scanf("%d",&n))
	{
	//	printf("%d:
",n);
		if(n<=3)
			printf("-1
");
		else if(n==4)
			printf("1 * 2
5 + 3
6 + 4
");
		else if(n==5)
			printf("1 * 2
3 / 6
4 - 7
5 * 8
");
		else if(n==6)
			printf("1 + 2
3 + 4
5 - 6
7 + 8
10 - 9
");
		else if(n==7)
			printf("1 + 2
3 + 8
9 / 4
10 + 5
11 + 6
12 + 7
");
		else if(n==8)
			printf("1 + 2
3 + 9
4 - 5
11 * 6
12 * 7
13 * 8
10 + 14
");
		else if(n==9)
			printf("1 + 2
3 + 10
4 / 5
6 / 7
8 / 9
11 - 12
15 - 13
 16 - 14
");
		else if(n==10)
			printf("1 + 2
3 / 4
5 / 6
7 / 8
9 / 10
11 + 12
16 + 13
17 + 14
18 + 15
");
		else if(n==11)
			printf("1 + 2
3 / 4
5 / 6
7 - 8
15 * 9
16 * 10
17 * 11
12 + 13
19 + 14
20 + 18
");
		else if(n==12)
			printf("1 + 2
3 - 4
5 * 14
6 * 15
7 * 16
8 * 17
9 * 18
10 * 19
11 * 20
12 * 21
13 + 22
");
		else if(n==13)
            printf("1 + 2
3 / 4
5 / 6
7 - 8
17 * 9
18 * 10
19 * 11
20 * 12
21 * 13
22 + 14
23 - 15
24 - 16
");
		else
		{
			printf("1 + 2
3 + 4
5 + 6
7 + 8
9 + 10
");
			printf("%d + %d
%d + %d
%d + %d
",n+1,n+2,n+3,n+4,n+5,n+6);
			printf("%d / 11
%d / 12
",n+7,n+8);
			printf("%d * %d
",n+9,n+10);
			printf("13 - 14
");
			tmp=n-14;
			int i;
			for(i=0;i<tmp;i++)
			{
				printf("%d * %d
",n+12+i,15+i);
			}
			printf("%d + %d
",n+11,n+12+tmp);
		}
	//	printf("
");
	}
	return 0;
}