DFS Additive equations （累加）

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

33 1 2 33 1 2 56 1 2 3 5 4 6

Output for the Sample Input

1+2=3Can't find any equations.1+2=31+3=41+4=51+5=62+3=52+4=61+2+3=6

题意：

给你一个集合，让你写出集合中满足的数学表达式，如果一个也没有输出  Can't find any equations.

思路：

从数学表达式的长度入手，从2到n-1，DFS
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int num[35],vis[35],n,flag;void DFS(int start,int len,int sum){    if(len==0)    {   //当长度为0说明表达式等号前的已经搜索完成        for(int i=start;i<=n&&sum>=num[i];i++)        { //从start开始 是有可能成为表达式等号后的数值，且等号前的和值要大于num[i]            if(sum==num[i])            {                flag=1;   //验证是否存在表达式                for(int j=1;j<=i;j++)                {  //表达式等号前的数字                    if(vis[j])                    {  //是否使用num[j]                      if(sum==num[j])                            printf("%d=%d\n",num[j],num[i]);                      else                             printf("%d+",num[j]);                        sum=sum-num[j];                    }                }            }        }    }    else    {        for(int i=start;i<=n;i++)        {              if(sum+num[i]<=num[n])            {  //表达式等号前的数值要小于集合种的最后一个数                sum=sum+num[i];  //计算和值                vis[i]=1;   //标记使用                --len;      //长度减一                DFS(i+1,len,sum);                vis[i]=0;                len++;                sum=sum-num[i];            }        }    }}int main(){    int icase;    scanf("%d",&icase);    while(icase--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)         scanf("%d",&num[i]);        memset(vis,0,sizeof(vis));        sort(num+1,num+n+1);        flag=0;        for(int i=2;i<n;i++)            DFS(1,i,0);        if(!flag)            printf("Can't find any equations.\n");            printf("\n");    }    return 0;}