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",n);
		if(n<=3)
			printf("-1\n");
		else if(n==4)
			printf("1 * 2\n5 + 3\n6 + 4\n");
		else if(n==5)
			printf("1 * 2\n3 / 6\n4 - 7\n5 * 8\n");
		else if(n==6)
			printf("1 + 2\n3 + 4\n5 - 6\n7 + 8\n10 - 9\n");
		else if(n==7)
			printf("1 + 2\n3 + 8\n9 / 4\n10 + 5\n11 + 6\n12 + 7\n");
		else if(n==8)
			printf("1 + 2\n3 + 9\n4 - 5\n11 * 6\n12 * 7\n13 * 8\n10 + 14\n");
		else if(n==9)
			printf("1 + 2\n3 + 10\n4 / 5\n6 / 7\n8 / 9\n11 - 12\n15 - 13\n 16 - 14\n");
		else if(n==10)
			printf("1 + 2\n3 / 4\n5 / 6\n7 / 8\n9 / 10\n11 + 12\n16 + 13\n17 + 14\n18 + 15\n");
		else if(n==11)
			printf("1 + 2\n3 / 4\n5 / 6\n7 - 8\n15 * 9\n16 * 10\n17 * 11\n12 + 13\n19 + 14\n20 + 18\n");
		else if(n==12)
			printf("1 + 2\n3 - 4\n5 * 14\n6 * 15\n7 * 16\n8 * 17\n9 * 18\n10 * 19\n11 * 20\n12 * 21\n13 + 22\n");
		else if(n==13)
            printf("1 + 2\n3 / 4\n5 / 6\n7 - 8\n17 * 9\n18 * 10\n19 * 11\n20 * 12\n21 * 13\n22 + 14\n23 - 15\n24 - 16\n");
		else
		{
			printf("1 + 2\n3 + 4\n5 + 6\n7 + 8\n9 + 10\n");
			printf("%d + %d\n%d + %d\n%d + %d\n",n+1,n+2,n+3,n+4,n+5,n+6);
			printf("%d / 11\n%d / 12\n",n+7,n+8);
			printf("%d * %d\n",n+9,n+10);
			printf("13 - 14\n");
			tmp=n-14;
			int i;
			for(i=0;i<tmp;i++)
			{
				printf("%d * %d\n",n+12+i,15+i);
			}
			printf("%d + %d\n",n+11,n+12+tmp);
		}
	//	printf("\n");
	}
	return 0;
}