Problem:
Problem F: Shredding Company 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.  The shredder takes as input a target number and a sheet of paper with a number written on it.  It shreds (or cuts) the sheet into pieces each of which has one or more digits on it.  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:  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.  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.  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:   t 1  num 1  t 2  num 2 ... t n  num n 0 0  Each test case consists of the following two positive integers, which are separated by one space: (1) the first integer ( t i above) is the target number; (2) the second integer ( num i 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  part 1  part 2 ...  rejected  error   In the first type, part j and sum have the following meaning:   Each part j is a number on one piece of shredded paper. The order of part j corresponds to the order of the original digits on the sheet of paper.   sum is the sum of the numbers after being shredded, i.e., sum = part 1 + part 2 + ... .   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  Output for the Sample Input  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

Solution (Python):


import sys
import math
import itertools as it
from collections import deque

sys.setrecursionlimit(10000000)

while True:
    N, S = raw_input().split()
    if N == "0" and S == "0":
        break
    N = int(N)
    m = {}
    def func(index, num, lst):
        if index == len(S):
            if num in m:
                m[num] = ["rejected"]
            else:
                m[num] = list(lst)
            return
        for i in range(index + 1, len(S) + 1):
            p = int(S[index:i])
            lst.append(str(p))
            func(i, num + p, lst)
            lst.pop()
    func(0, 0, [])
    score = -1
    ans = []
    for key in m:
        if score < key and key <= N:
            score = key
            ans = " ".join(m[key])
    if score == -1:
        print "error"
    elif ans == "rejected":
        print ans
    else:
        print str(score) + " " + ans
    pass

Solution (Python):
import itertools
while 1:
    t,s = map(int,raw_input().split())
    if t == 0: break
    ans = s if s <= t else 0
    s = str(s)
    nums = [s] if ans > 0 else []
    n = len(s)
    reject = False
    for split in xrange(1,n):
        for cuts in itertools.combinations(range(1,n),split):
            cuts = [0]+list(cuts)+[n]
            tmp = sum(int(s[i:j]) for i,j in zip(cuts,cuts[1:]))
            if   tmp == ans: reject = True
            elif t >= tmp > ans:
                ans = tmp
                nums = [s[i:j] for i,j in zip(cuts,cuts[1:])]
                reject = False
    if reject: print "rejected"
    elif len(nums) == 0: print "error"
    else: print ans, " ".join(nums)

Solution (Python):


def shred(num, bit):
    num = list(str(num))
    i = 0
    while bit:
        if bit & 1:
            num[i] += num[i+1]
            num.pop(i+1)
        else:
            i += 1

        bit >>= 1

    return num

def testcase_ends():
    t, num = map(int, raw_input().split())
    if t == 0 and num == 0:
        return 1

    cut = []
    best = -1
    for i in range(1<<(len(str(num))-1)):
        curcut = shred(num, i)
        cursum = sum(map(int, curcut))
        if cursum > t: continue
        if cursum == best:
            cut.append(curcut)
        elif cursum > best:
            cut = [curcut]
            best = cursum

    if not cut:
        print 'error'
    elif len(cut) > 1:
        print 'rejected'
    else:
        print best, ' '.join(cut[0])

def main():
    while True:
        if testcase_ends():
            return 0

if __name__ == '__main__':
    main()

Solution (Python):
def shred(num, bit):
    num = list(str(num))
    i = 0
    while bit:
        if bit & 1:
            num[i] += num.pop(i+1)
        else:
            i += 1

        bit >>= 1

    return num

def testcase_ends():
    t, num = map(int, raw_input().split())
    if t == 0 and num == 0:
        return 1

    cutcand = []
    curmax = -1
    for i in range(1<<(len(str(num))-1)):
        cut = shred(num, i)
        sumcut = sum(map(int, cut))

        if sumcut > t:
            continue
        elif sumcut == curmax:
            cutcand.append(cut)
        elif sumcut > curmax:
            cutcand = [cut]
            curmax = sumcut
    
    if len(cutcand) > 1:
        print 'rejected'
    elif len(cutcand) == 1:
        print curmax, ' '.join(cutcand[0])
    else:
        print 'error'

    return 0

def main():
    while True:
        if testcase_ends():
            return 0

if __name__ == '__main__':
    main()

Solution (Python):
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)


def main():
    rr = []
    ii = [2**i for i in range(6)]

    while True:
        t,n = LI()
        if t == 0 and n == 0:
            break

        s = [c for c in str(n)]
        l = len(s)

        r = None
        rm = -1
        rc = 0
        for i in range(2**(l-1)):
            k = []
            c = s[0]
            for j in range(l-1):
                if ii[j] & i:
                    c += s[j+1]
                else:
                    k.append(c)
                    c = s[j+1]
            k.append(c)
            m = sum(map(int,k))
            if m <= t:
                if rm < m:
                    rm = m
                    r = k
                    rc = 1
                elif rm == m:
                    rc += 1
        if rc == 0:
            rr.append('error')
        elif rc > 1:
            rr.append('rejected')
        else:
            rr.append('{} {}'.format(rm, ' '.join(r)))

    return '\n'.join(map(str, rr))


print(main())

Solution (Python):

while True:
    t, n = input().split()
    t = int(t)
    if t == 0:
        exit()
    L = len(n)
    m = 0
    parts = []
    f = False
    for b in range(1 << (L - 1)):
        c = int(n[0])
        s = 0
        li = []
        for k in range(L - 1):
            if b >> k & 1 == 1:
                s += c
                li.append(c)
                c = 0
            c = 10 * c + int(n[k + 1])
        s += c
        li.append(c)
        c = 0
        if s > t:
            continue
        if s >= m:
            f = (s == m)
            m = s
            parts = li
    if f:
        print('rejected')
    elif m == 0:
        print('error')
    else:
        print(m, *parts, sep=' ')

Solution (Python):
import itertools
from heapq import heappop, heappush
from collections import defaultdict
def main(t, num):
    num = list(str(num))
    d = defaultdict(int)
    if t < sum(map(int, num)):
        print("error")
        return
    q = [[0,[]]]
    itertools.product()
    fulls = itertools.product(range(2), repeat=len(num) - 1)
    for full in fulls:
        tmp = 0
        res = []
        b = 0
        f = False
        for i in range(len(num) - 1):
            if full[i]:
                res.append(i)
                tmp += int("".join(num[b: i + 1]))
                b = i + 1
        tmp += int("".join(num[b:]))
        a, b = heappop(q)
        a *= - 1
        if a == tmp: d[a] += 1
        if t >= tmp > a:
            a = tmp
            b = res
        heappush(q, [-a, b])
    a, b = heappop(q)
    a *= - 1
    if d[a] or a == 0:
        print("rejected")
    else:
        ans = []
        be = 0
        for i in b:
            ans.append(int("".join(num[be: i + 1])))
            be = i + 1
        ans.append(int("".join(num[be:])))
        print(a,*ans)

if __name__ == "__main__":
    while 1:
        t, num = map(int, input().split())
        if t == num == 0:
            break
        main(t, num)

Solution (Python):

from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve(t,n):
    if t == n:
        print(t,n)
        return
    a = list(map(int,list(str(n))))
    s = sum(a)
    m = len(a)
    if s > t:
        print("error")
        return
    nm = m-1
    f = [0]*nm
    d = defaultdict(lambda : 0)
    d[tuple(f)] = 1
    l = [[] for i in range(t+1)]
    q = [(s,f)]
    l[s].append(f)
    ans = s
    while q:
        s,f = heappop(q)
        for i in range(nm):
            if f[i]:
                continue
            nf = [1 if j == i else f[j] for j in range(nm)]
            ns = 0
            ms = a[0]
            for j in range(1,m):
                if nf[j-1]:
                    ms *= 10
                    ms += a[j]
                else:
                    ns += ms
                    ms = a[j]
            ns += ms
            if ns <= t:
                tf = tuple(nf)
                if not d[tf]:
                    d[tf] = 1
                    if ns >= ans:
                        ans = ns
                        l[ns].append(nf)
                    heappush(q,(ns,nf))
    if len(l[ans]) > 1:
        print("rejected")
        return
    s = [ans]
    f = l[ans][0]
    k = a[0]
    for i in range(nm):
        if f[i]:
            k *= 10
            k += a[i+1]
        else:
            s.append(k)
            k = a[i+1]
    s.append(k)
    print(*s)
    return


if __name__ == "__main__":
    while 1:
        t,n = LI()
        if t == n == 0:
            break
        solve(t,n)