Additive equations--zoj

Additive equations

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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We all understand that an integer set is a collection of distinct integers. Now the question is: given an integer set, can you find all its addtive equations? To explain what an additive equation is, let's look at the following examples: 

    1+2=3 is an additive equation of the set {1,2,3}, since all the numbers that are summed up in the left-hand-side of the equation, namely 1 and 2, belong to the same set as their sum 3 does. We consider 1+2=3 and 2+1=3 the same equation, and will always
output the numbers on the left-hand-side of the equation in ascending order. Therefore in this example, it is claimed that the set {1,2,3} has an unique additive equation 1+2=3.

    It is not guaranteed that any integer set has its only additive equation. For example, the set {1,2,5} has no addtive equation and the set {1,2,3,5,6} has more than one additive equations such as 1+2=3, 1+2+3=6, etc. When the number of integers in a set
gets large, it will eventually become impossible to find all the additive equations from the top of our minds -- unless you are John von Neumann maybe. So we need you to program the computer to solve this problem.

Input



The input data consists of several test cases. 

The first line of the input will contain an integer N, which is the number of test cases. 

Each test case will first contain an integer M (1<=M<=30), which is the number of integers in the set, and then is followed by M distinct positive integers in the same line.

Output



For each test case, you are supposed to output all the additive equations of the set. These equations will be sorted according to their lengths first( i.e, the number of integer being summed), and then the equations with the same length will be sorted according
to the numbers from left to right, just like the sample output shows. When there is no such equation, simply output "Can't find any equations." in a line. Print a blank line after each test case.

Sample Input

3
3 1 2 3
3 1 2 5
6 1 2 3 5 4 6

Output for the Sample Input

1+2=3

Can't find any equations.

1+2=3
1+3=4
1+4=5
1+5=6
2+3=5
2+4=6
1+2+3=6

第一个数字表示输入数据的数量，之后每行第一个数表示有几个可供使用的数字，然后用这些数字组成加法等式，同一个数字不能重复使用，但是多次输入的数字可以重复使用。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[50],b[50],vis[1000010],flog,n;
void bfs(int pos,int num,int sum,int key)
{/*pos指向a数组，选择a中的数，num表示b数组中数字个数，sum表示当前和
key表示最多使用的数字个数*/
	if(num>key)
	return ;
	if(sum>a[n-1])/*最大的和a数组的最后一个数*/
	return ;
	if(num==key&&vis[sum])
	{
		flog=0;/*当找到一种符合情况的时，flog赋值*/
		int i;
		for(i=0;i<key-1;i++)
		printf("%d+",b[i]);
		printf("%d=%d\n",b[key-1],sum);
		return ;
	}
	if(pos>=n)
	return ;
	b[num]=a[pos];
	bfs(pos+1,num+1,sum+a[pos],key);/*对于当前操作的数，有两种选择，
	要或不要，要的话num+1，否则不加，之后这个数会被下一个数覆盖*/
	bfs(pos+1,num,sum,key);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i;
		scanf("%d",&n);
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(vis,0,sizeof(vis));
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			vis[a[i]]=1;/*vis数组标记*/
		}
		flog=1;
		sort(a,a+n);
		for(i=2;i<n;i++)
		bfs(0,0,0,i);
		if(flog)
		printf("Can't find any equations.\n");
		printf("\n");
	}
	return 0;
}