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. 开发自己的 chart【转】

    Kubernetes 给我们提供了大量官方 chart,不过要部署微服务应用,还是需要开发自己的 chart,下面就来实践这个主题. 创建 chart 执行 helm create mychart 的 ...

  2. win10提示防火墙没有法更改某些设置的处理办法

    一.问题发现 远程链接电脑时间发现远程链接失败 提问在“控制面板” 中打开“程序” 列表中启用“windows 防火墙” . 按照提示启用防火墙 ,发现启用或关闭页面不可编辑 二.原因是防火墙Wind ...

  3. 剑指offer 把字符串转化为整数

    题目描述 将一个字符串转换成一个整数,要求不能使用字符串转换整数的库函数. 数值为0或者字符串不是一个合法的数值则返回0 输入描述: 输入一个字符串,包括数字字母符号,可以为空 输出描述: 如果是合法 ...

  4. ffmpeg 知识点

    ffmpeg FFmpeg是一套可以用来记录.转换数字音频.视频,并能将其转化为流的开源计算机程序.采用LGPL或GPL许可证.它提供了录制.转换以及流化音视频的完整解决方案.它包含了非常先进的音频/ ...

  5. 《YouTube 网站的架构演进》阅读笔记

    概述 YouTube 在国内是个404网站,需要翻墙得见,这是有用的废话,先铺垫一下. 从全球网站来看,它仅次于母公司 Google,全球排名位列第2.每天超过5亿以上视频播放量,平均每个用户点击10 ...

  6. ACM-生化武器

    Description在一个封闭的房间里,gogo给大家表演了他的屁遁术,人果然一下没影了,但是他留下的“生化武器”,却以每秒1米的速度向上下左右扩散出去.为了知道自己会不会被“毒”到,你果断写了个算 ...

  7. 从高通出重拳,iPhone遭禁售看巨头们的专利大战

    一直以来,高通看起来都是很有"骨气"的.面对与苹果的专利纷争,高通始终是据理力争.顽强对抗.截至今年6月中旬,高通和苹果打了50多场专利官司,且这些官司分布于6个不同国家的16个司 ...

  8. JS的数据类型、常量、变量、以及基本对象的知识总结

    第一部分.JS的常见数据类型,特别要注意的是JS中大小写要求很严格,一定要注意字段大小写. 1.字符串(String) 举例: var cellname = "Bill Gate" ...

  9. LCT(2)

    LCT(2) 关于 LCT 的基本操作和代码实现见 (1) . 5. LCT的应用 5.0 LCT 裸题 就是LCT的基本操作模板题,常出现于早年省选.不讨论. 5.1 LCT维护子树信息 很多时候, ...

  10. 自定义jqGrid编辑功能,当行获取焦点时编辑,失去焦点时保存

    http://www.360doc.com/content/17/0719/15/9200790_672577533.shtml /********************************** ...