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.

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 x 1 , . . . , x n . 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 x 1 , . . . , x n 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.

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.

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

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

题意：
   在n个数中选出几个数，使它们相加等于t；并按照从大到小的顺序输出这些数。

题解：
   dfs水题，就算水题，我也做了好一会，太菜了。这道题很显然，因为题中给出的顺序就是从大到小的，所以我们从第一个数开始，依次往后搜索，将可能的数据都记录下来，每遇到一种满足题意的组合就输出，一直搜索下去，得到所有答案。若没有答案，输出NONE。

代码：

#include <iostream>

#include <stdio.h>

#include <algorithm>

using namespace std;

int n,t,a[],f=,b[];

void dfs(int len,int k,int sum)

{

    if(sum==){    //遇到满足题意的，输出。

        f=;

        printf("%d",b[]);

        for(int i=;i<len;i++)

            printf("+%d",b[i]);

        printf("\n");

        return ;

    }

    for(int i=k;i<n;i++){

        if((i==k||a[i]!=a[i-])&&sum-a[i]>=){  //防止重复

            b[len]=a[i];

            dfs(len+,i+,sum-a[i]);

        }

    }

}

int main()

{

    while(){

        cin>>t>>n;

        f=;

        if(t==&&n==) break;

        for(int i=;i<n;i++) cin>>a[i];

        printf("Sums of %d:\n",t);

        dfs(,,t);

        if(f) printf("NONE\n");

    }

    return ;

}