ZOJ 1204 一个集合能组成多少个等式

Additive equations

Time Limit : 20000/10000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 22   Accepted Submission(s) : 12
[align=left]Problem Description[/align]
    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 anadditive 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

 

 

[align=left]Source[/align]
Zhejiang University Local Contest 2002, Preliminary
 
题意：  输入一个集合 看这个集合中能组成多少个等式    注意1+2=3和2+1=3一样   
另外输出首先按照参与元素个数多少排序    如果参与个数相同 则按照元素大小排序 如上面样例
 
思路：
DFS 
DFS 搜索
 

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<string>
using namespace std;
int st[100],n,vis[100];
string str;
string ans[33][6000];
int a[33];
int cnt=0;
void change(int mid)
{
int num=0;
char ss[22];
while(mid)
{
ss[num++]=mid%10+'0';
mid=mid/10;
}
for(int i=num-1;i>=0;i--) str+=ss[i];
}
void out()
{
int i,flag=1;
cnt=0;
str.clear();
for(i=0;i<n;i++)
{
if(vis[i])
{
if(flag==0)
str+='+';
flag=0;
change(st[i]);
cnt++;
}
}
}
void DFS(int sum,int pos)
{
int i;
if(sum>st[n-1]) return ;
for(i=pos+1;i<n;i++)
{
if(sum==st[i])
{
out();
str+='=';
change(st[i]);
cnt++;
ans[cnt][a[cnt]++]=str;
}
vis[i]=1;
DFS(sum+st[i],i);
vis[i]=0;
}
}
int main()
{
int cas,i;
scanf("%d",&cas);
while(cas--)
{
scanf("%d",&n);
for(i=0;i<n;i++)  scanf("%d",&st[i]);
sort(st,st+n);
memset(a,0,sizeof(a));
memset(vis,0,sizeof(vis));
for(i=0;i<n;i++)
{
vis[i]=1;
DFS(st[i],i);
vis[i]=0;
}
int ok=0;
for(i=3;i<33;i++)
{
for(int j=0;j<a[i];j++)
{
ok=1;
printf("%s\n",ans[i][j].c_str());
}
}
if(ok==0)
{
printf("Can't find any equations.\n");
}
printf("\n");
}
return 0;
}