Sum It Up

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 6

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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

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

這道題的關鍵就是去重了因為有重複數字嘛，比如：400 12 50 50 50 50 50 50 25 25 25 25 25 25這個例子有重複數字，如果全排列的話會重複，所以去重就好了，最簡單的一種就是開一個數組，然後重複的不忘進去放，但是既然是遞迴的話我們就可以利用遞迴的特性，我設一個標記位，把上一位的標記如果我當前這個大於這個標記位才可以往進去放。所以就完美解決了這個不重複全排列，很水的一道題。

# include <iostream>
# include <cstring>

using namespace std;

const int maxn = 50;
int a[maxn];
int b[maxn];
int sum, t;
bool flag;

void dfs(int vis, int f, int s) //vis是下標,f是當前b陣列的下標，s是和
{
   if(s > sum) return ;
   if(s == sum)
   {
       flag = true;
     for(int i = 0; i < f; i++)
       {
           i != f - 1? cout << b[i]<< "+" : cout << b[i] << endl;
       }
   }
   int k = -1;
   for(int i = vis; i < t; i++)
   {
       if(k != a[i])
       {
           b[f] = a[i];
           k = a[i];
           dfs(i + 1, f + 1, s + b[f]);
       }
   }
}

int main(int argc, char *argv[])
{
    while(cin >> sum >> t, t || sum)
    {
        for(int i = 0; i < t; i++)
        {
            cin >> a[i];
        }
        cout << "Sums of "<< sum <<":" << endl;
        flag = false;
        dfs(0, 0, 0);
        if(!flag)
        {
            cout << "NONE" << endl;
        }
    }
    return 0;
}