Shredding Company

Problem Description

You have just been put in charge of developing a new shredder for the Shredding Company Although a "normal" shredder would just shred sheets of paper into little pieces so that the contents would become unreadable, this new shredder needs to have the following unusual basic characteristics. 

1.The shredder takes as input a target number and a sheet of paper with a number written on it. 

2.It shreds (or cuts) the sheet into pieces each of which has one or more digits on it. 

3.The sum of the numbers written on each piece is the closest possible number to the target number, without going over it. 

For example, suppose that the target number is 50, and the sheet of paper has the number 12346. The shredder would cut the sheet into four pieces, where one piece has 1, another has 2, the third has 34, and the fourth has 6. This is because their sum 43 (= 1 + 2 + 34 + 6) is closest to the target number 50 of all possible combinations without going over 50. For example, a combination where the pieces are 1, 23, 4, and 6 is not valid, because the sum of this combination 34 (= 1 + 23 + 4 + 6) is less than the above combination's 43. The combination of 12, 34, and 6 is not valid either, because the sum 52 (= 12 + 34 + 6) is greater than the target number of 50. 
 
Figure 1. Shredding a sheet of paper having the number 12346 when the target number is 50

There are also three special rules : 

1.If the target number is the same as the number on the sheet of paper, then the paper is not cut. 

For example, if the target number is 100 and the number on the sheet of paper is also 100, then 

the paper is not cut. 

2.If it is not possible to make any combination whose sum is less than or equal to the target number, then error is printed on a display. For example, if the target number is 1 and the number on the sheet of paper is 123, it is not possible to make any valid combination, as the combination with the smallest possible sum is 1, 2, 3. The sum for this combination is 6, which is greater than the target number, and thus error is printed. 

3.If there is more than one possible combination where the sum is closest to the target number without going over it, then rejected is printed on a display. For example, if the target number is 15, and the number on the sheet of paper is 111, then there are two possible combinations with the highest possible sum of 12: (a) 1 and 11 and (b) 11 and 1; thus rejected is printed. In order to develop such a shredder, you have decided to first make a simple program that would simulate the above characteristics and rules. Given two numbers, where the first is the target number and the second is the number on the sheet of paper to be shredded, you need to figure out how the shredder should "cut up" the second number. 

 

Input

The input consists of several test cases, each on one line, as follows :

tl num1
t2 num2
...
tn numn
0 0

Each test case consists of the following two positive integers, which are separated by one space : (1) the first integer (ti above) is the target number, (2) the second integer (numi above) is the number that is on the paper to be shredded.

Neither integers may have a 0 as the first digit, e.g., 123 is allowed but 0123 is not. You may assume that both integers are at most 6 digits in length. A line consisting of two zeros signals the end of the input.

 

Output

For each test case in the input, the corresponding output takes one of the following three types :

sum part1 part2 ...
rejected
error

In the first type, partj and sum have the following meaning :

1.Each partj is a number on one piece of shredded paper. The order of partj corresponds to the order of the original digits on the sheet of paper.

2.sum is the sum of the numbers after being shredded, i.e., sum = part1 + part2 +...

Each number should be separated by one space.
The message error is printed if it is not possible to make any combination, and rejected if there is
more than one possible combination.
No extra characters including spaces are allowed at the beginning of each line, nor at the end of each line.

 

Sample Input

50 12346 376 144139 927438 927438 18 3312 9 3142 25 1299 111 33333 103 862150 6 1104 0 0

 

Sample Output

43 1 2 34 6 283 144 139 927438 927438 18 3 3 12 error 21 1 2 9 9 rejected 103 86 2 15 0 rejected

 

      说是有一个碎纸机，把一个有一串数的纸条放进去，可以把纸条截开成几部分，每一部分成为一个数，求把这些数加起来的和sum，和一个已给出的数x比较。如果没有小于x的sum，输出“error”；如果存在小于x的sum，且其中最大的sum只出现一次，输出sum和每个部分的数；如果这个最大的sum出现两次及以上，输出“rejected”。

     理解了之后，想法是这样的：比方一串abcde，先看a与b，a可能是单独被分成一部分，b也单独成为一部分；也可能a和b一起，成为（a*10+b）；不管变成哪一个，假设为temp，要和c进行组合了，也是两种情况：1.temp、c两部分   2.（temp*10+c） 。根据这个进行搜索就可以了。

    下面是我写的两个代码，基本想法是一样的，按照上面方法搜索，但是一个对的一个错的，我始终没找出来原因。

AC代码：

#include
#include
#include
#include
using namespace std;
int num,len,ans,x,reject;
int p[10];
int a[10];
int answer[10];
void ioan(int sum,int now,int k,int t)
{
    if(sum>num)
        return ;
    if(k==len)
    {
        sum+=now;
        a[t]=now;
        if(sum>num)
            return ;
        if(ans<=sum)
        {
                if(ans>num)
    {
        cin>>str;
        if(num==0 && str=="0")
            break;
        len=str.size();
        reject=0;
        ans=0;
        x=0;
        memset(p,0,sizeof(p));
        memset(a,0,sizeof(a));
        memset(answer,0,sizeof(answer));
        int s=0;
        for(int i=1;i<=len;i++)
        {
            p[i]=str[i-1]-'0';
            s+=p[i];
        }
        if(s>num)
        {
            cout<<"error"<





未AC代码：

#include
#include
#include
#include
using namespace std;
int num,len,xx,ans,reject;
int p[20];
int a[20];
int aa[1000000];
int answer[20];
//sum:全部总和  now:目前判断数  k:在第几个数上  t:存了几个数了
void ioan(int sum,int now,int k,int t)
{
    if(sum>num)
        return ;
    if(k==len)
    {
        if(ans<=sum && sum<=num)
        {
            if(ans>num)
    {
        cin>>str;
        if(num==0 && str=="0")
            break;
        len=str.size();
        xx=0;
        ans=0;
        reject=0;
        int s=0;
        memset(aa,0,sizeof(aa));
        memset(p,0,sizeof(p));
        memset(answer,0,sizeof(answer));
        for(int i=1;i<=len;i++)
        {
            p[i]=str[i-1]-'0';
            s+=p[i];
        }
        if(s>num)
        {
            cout<<"error"<