Given a specifie d total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n = 6, and the list is [4, 3, 2, 2, 1, 1], then there are four different sums that equal 4: 4, 3+1, 2+2, and 2+1+1. (A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.
Input
The input file will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers 
. If n = 0 it signals the end of the input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12 (inclusive), and will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear in nonincreasing order, and there may be repetitions.
Output
For each test case, first output a line containing `Sums of ', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line `NONE'. The numbers within each sum must appear in nonincreasing order. A number may be repeated in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distinct; the same sum cannot appear twice.
Sample Input
4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0
Sample Output
Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25
#include<iostream>
#include<cstdlib>
#include<memory>
using namespace std;
int mount;
int l;
int a[12][2];
int mark;
void output()
{
mark++;
int temp[12];
int flag=0;
for(int i=0;i<mount;i++)
{
if(a[i][1])
{
if(flag==0)
flag++;
else
cout<<"+";
cout<<a[i][0];
}
}
cout<<endl;
}
void dfs(int sum,int nowv,int pre)
{
int value;
if(nowv==sum)
{
output();
}
else
{
for(int j=pre+1;j<mount;j++)
{
value=a[j][0];
if(nowv+value<=sum&&value!=l)
{
a[j][1]=1;
dfs(sum,nowv+value,j);
a[j][1]=0;
}
}
}
if(pre!=-1)
l=a[pre][0];
}
int main()
{
//freopen("s.txt","r",stdin);
//freopen("key.txt","w",stdout);
int sum,i,j;
cin>>sum>>mount;
while(sum!=0)
{
for(i=0;i<mount;i++)
{
cin>>a[i][0];
a[i][1]=0;
}
mark=0;
l=0;
cout<<"Sums of "<<sum<<":"<<endl;
dfs(sum,0,-1);
if(mark==0)
cout<<"NONE"<<endl;
cin>>sum>>mount;
}
//system("PAUSE");
return 0;
}
難點(diǎn)：1，有序遞減，這個(gè)好處理，只需在向前推時(shí)dfs(i)for(int j=i+1...)即可
2，去掉無(wú)效的重復(fù)搜索，這個(gè)煩一點(diǎn),比如和為11時(shí)，序列為6,6,6,2,2,2,1，6+2+2+1顯然是可以的，但是6+6不可以。6+6+6更不可以
我們可以這樣想，如果順著往前推是，如果不超過(guò)范圍，比如6+2+取后一個(gè)數(shù)時(shí)，2可以取，但是6+6，超過(guò)11，要回溯的時(shí)候，記錄下此時(shí)的6，用下面的代碼
if(pre!=-1)
l=a[pre][0];
在比較if(nowv+value<=sum&&value!=l)此時(shí)不符合，下一個(gè)6取不到了！避免了重復(fù)。