Shredding Company
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 4713   Accepted: 2714

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

题意是给定一个最多是六位数的纸片,然后给了一个target数字,对这个纸片进行切割得到不同的数字,在这些数字之和 中找到小于等于target的最大数字。

还有一些其他的小规则:

切割的时候,如果该纸片上的数字是target,该纸片不能切割。

如果这个纸片无论怎么切割,数字之和都大于target,输出error。

如果最佳方案有两种或两种以上,输出rejected。

排除以上情况,输出最优结果,以及在最优结果下的切割方案。

又是一道折磨我许久的题。。。

最优结果能求出,但最佳方案不会弄出来,结果还是发现在dfs中传递数组,见识了不同的dfs,真的是无所不能。这几天被挫的厉害啊。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std; int target, len, maxn, num_max;//数组的最大长度
int val_final[8], val[8];
int dis[1000000];
string shre; void dfs(int x, int sum, int c)
{
if (x > len - 1)
{
if (maxn < sum)
{
maxn = sum;
num_max = c;
for (int i = 0; i < c; i++)
{
val_final[i] = val[i];
}
}
dis[sum]++;
return;
}
else
{
int i, t = 0;
for (i = x; i < len; i++)
{
t = t * 10 + shre[i] - '0';
if (sum + t <= target)
{
val[c] = t;
dfs(i + 1, sum + t, c + 1);
}
else
break;
}
}
} int main()
{
int i, t;
while (cin>>target>>shre)
{
if (target == 0 && shre == "0")
break;
memset(dis, 0, sizeof(dis));
memset(val_final, 0, sizeof(val_final));
memset(val, 0, sizeof(val));
t = 0;
len = shre.length();
maxn = -1; for (i = 0; i < len; i++)
{
t += shre[i] - '0';
}
if (t > target)
{
cout << "error" << endl;
continue;
} dfs(0, 0, 0); if (dis[maxn] != 1)
{
cout << "rejected" << endl;
}
else
{
cout << maxn;
int k;
for (k = 0; k < num_max; k++)
{
cout << " " << val_final[k];
}
cout << endl;
}
}
return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

POJ 1416:Shredding Company的更多相关文章

  1. 【POJ 1416】Shredding Company

    题 题意 给你一个target number,和一个最多六位的数num,让你把数分段,使总和最接近但不大于target number. 如果只有一种方法就输出总和.分段,如果有多种方法,输出rejec ...

  2. POJ 1416 Shredding Company【dfs入门】

    题目传送门:http://poj.org/problem?id=1416 Shredding Company Time Limit: 1000MS   Memory Limit: 10000K Tot ...

  3. 搜索+剪枝 POJ 1416 Shredding Company

    POJ 1416 Shredding Company Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5231   Accep ...

  4. POJ 1416 Shredding Company 回溯搜索 DFS

    Shredding Company Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6173   Accepted: 3361 ...

  5. poj1416 Shredding Company

    Shredding Company Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5379   Accepted: 3023 ...

  6. POJ1416——Shredding Company(DFS)

    Shredding Company DescriptionYou have just been put in charge of developing a new shredder for the S ...

  7. Shredding Company(dfs)

    Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3519   Accepted: 2009 Description You h ...

  8. Shredding Company (hdu 1539 dfs)

    Shredding Company Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others ...

  9. 【POJ1416】Shredding Company

    本题传送门 本题知识点:深度优先搜索 + 回溯 本题题意很简单,就是有一条位数不超过6的数字纸条,问你怎么剪这纸条,使得得到的纸条的值的总和最接近目标值(总和不能超过目标值). 比如第一个样例 50 ...

随机推荐

  1. 第四张5G牌照发给广电,能打破三大运营商的垄断吗?

    近段时间,多个国家处于莫须有的安全性考虑,禁止华为参与核心5G网络设备竞标.其实这就从侧面反映出,国内民族企业在5G层面的领先性.当然,这也让我们认知到,5G网络将是新时代的竞争关键节点.为此,国内正 ...

  2. Redis详解(八)——企业级解决方案

    Redis详解(八)--企业级解决方案 缓存预热 缓存预热就是系统上线后,提前将相关的缓存数据直接加载到缓存系统.避免在用户请求的时候,先查询数据库,然后再将数据缓存的问题!用户直接查询事先被预热的缓 ...

  3. IDEA中利用MAVEN制作和打包普通可执行应用(非SprintBoot的WEB应用)

    我使用IDEA也就半年,开发中常常会遇到一些问题,例如用IDEA编写普通的可执行程序: 之前使用Eclipse编写一个可执行的JAVA程序,然后导出打包,非常方便: 但是我呢,想在 IDEA 中用Ma ...

  4. 053、Java中使用for循环实现1~100的累加

    01.代码如下: package TIANPAN; /** * 此处为文档注释 * * @author 田攀 微信382477247 */ public class TestDemo { public ...

  5. 删除C代码中的注释行【状态机】

    今天在学ruby时遇到的一个经典的题目,一直都知道但从来没有实现过.呈上状态机,代码略.(写代码的时候还是需要注意一些小情况的)

  6. 如何让你的学术Essay写作看起来更高级?

    学术风格不仅是使用某些特定的词汇或者特定的表达方式.学术风格还会涉及到你的思考方式.推理方式以及你如何利用某个领域内已知的发现.学术风格是英国大学学习的核心.所以在平时的学习中,英国环球论文小编建议大 ...

  7. ConfigureDefender – Windows Defender 设置工具

    用于配置Windows 10内置Defender防病毒设置的实用程序.它是一个便携式实用程序 ConfigureDefender实用程序是一个GUI应用程序,用于在Windows 10上查看和配置重要 ...

  8. (最详细)JAVA如何连接虚拟机的HBASE和hadoop(JAVA如何远程访问虚拟机HBASE)

    第一步: 首先把虚拟机和你的主机(本地电脑)弄通这样本地机器才能访问虚拟机里面的内容 我用的虚拟机为 VMware Workstation linux 为 centeros 补充一点虚拟机设置 1  ...

  9. Django(八)模型:Mysql8.0日志文件配置开启/关闭、查询

    Mysql日志文件开启.配置.查看 mysql.log是mysql的日志文件,里面记录的对MySQL数据库的操作记录.默认情况下mysql的日志文件没有产生,需要修改mysql的配置文件,步骤如下: ...

  10. maven项目添加配置文件

    1. 在src/main/resources下新建param.properties 2. 在param.properties文件中添加 mqtt.host=tcp://127.0.0.1:1883 m ...