设为首页 加入收藏

TOP

poj 1416 Shredding Company dfs(二)
2015-07-20 17:43:52 来源: 作者: 【 】 浏览:7
Tags:poj 1416 Shredding Company dfs
n 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

题意及分析:

题意不是很麻烦的那种。就不多说了。题目出了后就觉得这个题也没什么好说的。哎,上代码吧。

AC代码:

首页 上一页 1 2 下一页 尾页 2/2/2
】【打印繁体】【投稿】【收藏】 【推荐】【举报】【评论】 【关闭】 【返回顶部
分享到: 
上一篇Leetcode dfs Path SumII 下一篇Leetcode dp Word Break

评论

帐  号: 密码: (新用户注册)
验 证 码:
表  情:
内  容:

·C++中智能指针的性能 (2025-12-25 03:49:29)
·如何用智能指针实现c (2025-12-25 03:49:27)
·如何在 C 语言中管理 (2025-12-25 03:20:14)
·C语言和内存管理有什 (2025-12-25 03:20:11)
·为什么C语言从不被淘 (2025-12-25 03:20:08)