1197: Sum It Up

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本文探讨了一种经典的组合问题——寻找所有可能的数的组合，这些数来自一个指定列表并加起来等于一个特定的总和。文章详细介绍了如何通过递归深度优先搜索算法来解决该问题，并给出了具体的实现代码及样例输入输出。

1197: Sum It Up

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	624	242	Standard

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 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

This problem is used for contest: 60 149

#include<iostream>
#include<algorithm>
using namespace std;
int c[100];
bool flag ;
int ans[100];
int t,n;
int len;
bool cmp(int a,int b)
{
    return a>b;
}
void out()
{
    int j;
printf("%d", ans[0]);
    for (j = 1; j < len; j++)
            printf("+%d", ans[j]);
        printf("/n");
}
void dfs(int sum,int mark)
{
    int i;
if(sum==0)
    {
  flag=true;
  out();
  return ;
}
if(sum<0||mark>=n) return ;
for(i=mark;i<n;i++)
{
  if(i==mark||c[i]!=c[i-1])
  {
   ans[len++]=c[i];
   dfs(sum-c[i],i+1);
   len--;
  }
}
}
int main()
{
    int i;
    while(scanf("%d%d",&t,&n)&&t&&n)
    {
        for(i=0;i<n;i++)
        {
            cin>>c[i];
        }
        sort(c,c+n,cmp);
  printf("Sums of %d:/n", t);
         flag=false ;
   len=0;
         dfs(t,0);
         if(flag==false) cout<<"NONE"<<endl;
    }
    return 0;
}
// 6 5 5 2 2 2 1
//400 12 50 5 51 50 5 59 25 25 5 25 5 25