POJ 1416 Shredding Company__深搜

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.

1. 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.

2. 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

分析

题目大意是给出两个数字，后一个数字分割相加的和需要小于或等于前一个数字，前满足条件的最大值。在这边我是从左到右找所有分割组合，下面的代码注释很详细。郁闷的是先开始把continue写成return了竟然测试用例全过了，后来测试10000/10111才发现问题，醉。。。

实现

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;
#define NOT_PATH -1
const string REJECTED = "rejected";
const string ERROR = "error";
int maxSum, maxSumCount, step;
int a[16], result[16];

void dfs(int restTarget, int restNum, int numDigit, int prevSum)
{
    if (restTarget < 0) return;
    if (numDigit <= 0) return;

    //最大分割位数
    int targetDigit = restTarget > 0 ? int(log10(double(restTarget))) + 1 : 1;
    //实际位数不相等说明restNum前面有0。
    int realNumDigit = restNum > 0 ? int(log10(double(restNum))) + 1 : 1;
    int headZeroDigit = numDigit - realNumDigit;
    //先分割成大数字，再分割成小数字，减少分割的次数，加headZeroDigit只是为了记录分割方案。
    for (int i = targetDigit + headZeroDigit; i >= 1; i--) {
        if (i > numDigit) return;
        int tailNumberDigit = numDigit - i;
        int factor = pow(double(10), tailNumberDigit);
        //分割的前半段
        int headNumber = restNum / factor;
        //分割的后半段
        int tailNumber = restNum % factor;
        //找到符合条件的分割点则计算sum.
        if ((headNumber + tailNumber) <= restTarget) {
            //不超过要拼装的数则检查最大和是否需要更新，同时后续即使再分割也不会有比currentSum更大的数，不必再继续搜索。
            int currentSum = prevSum + headNumber + tailNumber;
            if (maxSum <= currentSum) {
                maxSumCount = (maxSum < currentSum) ? 1 : maxSumCount + 1;
                maxSum = currentSum;
                memcpy(result, a, sizeof(result));
                //附上最后分割的两个数字
                int resultIndex = step;
                if (headZeroDigit > 0) {
                    for (int j = 0; j < headZeroDigit; j++) {
                        result[resultIndex++] = 0;
                    }
                    //出现0表示能分割成的方案数肯定超过一个，设个大数供打印处判断。
                    maxSumCount = 1 << 30;
                }
                result[resultIndex++] = headNumber;
                int tailZeroDigit = (numDigit - i) - (tailNumber > 0 ? int(log10(double(tailNumber))) + 1 : 1);
                if (tailZeroDigit > 0) {
                    for (int j = 0; j < tailZeroDigit; j++) {
                        result[resultIndex++] = 0;
                    }
                    //出现0表示能分割成的方案数肯定超过一个，设个大数供打印处判断。
                    maxSumCount = 1 << 30;
                }
                if (tailNumberDigit != 0) {
                    result[resultIndex] = tailNumber;
                }
            }
            continue;
        }
        a[step++] = headNumber;
        dfs(restTarget - headNumber, tailNumber, tailNumberDigit, prevSum + headNumber);
        a[--step] = NOT_PATH;
    }
}

int main()
{
//  freopen("in.txt", "r", stdin);
    int target, num;
    while (cin >> target >> num && !(target == 0 && num == 0)) {
        maxSum = maxSumCount = step = 0;
        memset(a, NOT_PATH, sizeof(a));
        memset(result, NOT_PATH, sizeof(result));
        int numDigit = num > 0 ? int(log10(double(num))) + 1 : 1;
        dfs(target, num, numDigit, 0);
        if (maxSumCount == 0) {
            cout << ERROR << endl;
        }
        else if (maxSumCount > 1) {
            cout << REJECTED << endl;
        }
        else {
            cout << maxSum;
            for (int i = 0; i < 6; i++) {
                if (result[i] != NOT_PATH) {
                    cout << " " << result[i];
                }
            }
            cout << endl;
        }
    }
    return 0;
}