传送门：
http://acm.hdu.edu.cn/showproblem.php?pid=1258
Sum It Up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7758 Accepted Submission(s): 4067
Problem Description
Given a specified 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 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 x1,...,xn. 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 x1,...,xn 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 distince; the same sum connot 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
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
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
Source
Recommend
题目意思：
给定一个非递减的序列，要求从这些序列中找出一系列的数相加等于要求的数。
主要就是我们需要对结果去重！
这里有一共很巧妙的办法
当你要搜的数字刚好前一步搜了，那么跳过你要搜的数字（直接看代码更好理解）
还有就是这个dfs的框架也很重要
关于到它到底是怎么搜的
我觉得很受启发
code：
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<memory.h>
using namespace std;
#define max_v 25
int sum,n;
int a[max_v];
int ans[max_v];
int flag;
bool cmp(int x,int y)
{
return x>y;
}
void dfs(int now,int s,int f)//现在的和 开始位置 结束位置
{
if(now>sum)//没有搜到
return ;
if(now==sum)//搜到 输出
{
cout<<ans[];
for(int i=;i<f;i++)
{
cout<<"+"<<ans[i];
}
cout<<endl;
flag=;
}
for(int i=s;i<n;i++)
{
ans[f]=a[i];
dfs(now+a[i],i+,f+);
while(i+<n&&a[i]==a[i+])//搜完后，若下一个要搜的和搜完的一样 跳过
{
i++;
}
}
}
int main()
{
while(cin>>sum>>n)
{
if(sum==&&n==)
break;
int allsum=;
for(int i=;i<n;i++)
{
cin>>a[i];
allsum+=a[i];
}
cout<<"Sums of "<<sum<<":"<<endl;
if(allsum<sum)
{
cout<<"NONE"<<endl;
continue;
}
sort(a,a+n,cmp);
flag=;
dfs(,,);
if(flag==)
{
cout<<"NONE"<<endl;
}
}
return ;
}