poj1416 Shredding Company

Shredding Company Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5381 Accepted: 3023

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

题目翻译:

公司现在要发明一种新的碎纸机，要求新的碎纸机能够把纸条上的数字切成最接近而不超过target值。比如，target的值是50，而纸条上的数字是12346，应该把数字切成四部分，分别是1、2、34、6。因为这样所得到的和43 (= 1 + 2 + 34 + 6) 是所有可能中最接近而不超过50的。（比如1, 23, 4, 和6 就不可以,因为它们的和不如43接近50，而12, 34, 6也不可以，因为它们的和超过50了。碎纸还有以下三个要求：

1、如果target的值等于纸条上的值，则不能切。 2、如果没有办法把纸条上的数字切成小于target，则输出error。如target是1而纸条上的数字是123，则无论你如何切得到的和都比1大。 3、如果有超过一种以上的切法得到最佳值，则输出rejected。如target为15，纸条上的数字是111，则有以下两种切法11、1或者1、11. 你的任务是编写程序对数字进行划分以达到最佳值。（也就是能够获得的小于目标值的最大值）

人家说的可是比我明白多了，直接拿来用吧qwq

看，就是这里的：http://user.qzone.qq.com/289065406/blog/1304031265

当然我表示只是去看一下具体怎么穷举所有情况哒，我的代码要棒多了呐(=@__@=)（啪）

解题思路： 用DFS深搜穷举所有情况

(1)比如一个6位数n，切成为6个数的话，这6个数的和如果大于目标数goal则不用再搜索了，因为这肯定是所有划分中和最小的，而最小都比目标数大，自然就没有合要求的答案了. (2) 如何切分，假如以50  12346 为例。 第一步，先切下一个“1”，然后递归去切“2346”； 第二步，再切下一个“12”，然后递归去切“346”； 第三步，再切下一个“123”，然后递归去切“46”； 第四步，再切下一个“1234” 然后递归去切“6”  第五步，再切下“12346”。

（这样做的确是穷举了所有情况，又学习了qwq）

(3)切下来的 前面的数字串部分 则加入到划分的和，剩下的部分继续递归，直到剩下的数字串长度为0。 我直接用一个串记录划分方式( p)， 如上例的输入为50  12346时，其结果为43  1  2  34  6，那么p=1121，代表把12346划分为4部分，第一部分为第1位，第二部分为第2位，第三部分为第3、4位，第四部分为第5位

(4)注意在搜索时，穷举的时候要直接对字符串进行处理再搜索，不然若 剩余的数字开头第一位为 0 时，会导致出错。譬如：7 1023 实际上是有多重分割方式 1 0 2 3和1 02 3，qwq

(5)剪枝方法：在搜索时若发现部分和 大于（因为再往下会更大，不可能等于goal了）goal时，则可结束搜索。

(6)error的判定要在搜索前进行（分好情况就可以辣），rejected（多个最优解）的判定要在搜索后判定。

(7)关于出现相同最优解的标记，每出每种划分的sum每出现一次标记+1，要使标记为O(1)，只需把vist数组开得足够大。N最多为6位数，完全没压力～

#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <algorithm> #include <queue> #include <vector> #include <map> using namespace std; const int MAXN=1000000+5; int goal,sum,ans,cnt;//goal是目标数，sum用来累加切分的和，ans是最终答案，cnt是p的计数器 char p[10],path[10];//p用来临时记录路径，path用来记录最终路径 char num[10];//这是纸qwq int vis[MAXN];//记录答案个数 int getnum(char *s,int l)//这个函数就是把串换成整数 { int i; int sum=s[0]-'0'; for(i=1;i<l;++i)sum=sum*10+s[i]-'0'; return sum; } void dfs(char *s,int len) { int i; if(!len)//不能再切割了，就更新状态 { vis[sum]++; if(sum>ans&&sum<=goal) { ans=sum; p[cnt]=0; strcpy(path,p); } return ; } for(i=1;i<=len;++i)//穷举1-len位可以切割的情况 { int a=getnum(s,i); sum+=a; if(sum>goal)//如果此时的和就大于goal那么没必要继续搜索了 { sum-=a; //直接回溯 return ; } p[cnt++]=i; //记录路径 dfs(s+i,len-i);//继续递归剪切后面的串 sum-=a; //回溯 cnt--; } } int main() { int i,j; while(~scanf("%d %s",&goal,num)) { if(!goal&&num[0]=='0')break; sum=0; int len=strlen(num); if(getnum(num,len)==goal)//如果两个数相同，那么直接输出就可以了，因为搜索的数只能比它小 { printf("%d %d\n",goal,goal); continue; } //否则的话看一下最小的数，是否比goal大 for(i=0;i<len;++i)sum+=num[i]-'0'; if(sum>goal) { printf("error\n"); continue; } //如果都不符合，开始搜索 sum=0; cnt=0; ans=-1; memset(vis,0,sizeof(vis));//初始化状态 QAQ忘清零了 dfs(num,len); if(vis[ans]>1) { printf("rejected\n");//如果两次都搜到了，说明存在多种情况 } else { printf("%d",ans); int l=strlen(path); int top=0; for(i=0;i<l;++i) { printf(" "); for(j=0;j<path[i];++j)printf("%c",num[top++]); } printf("\n"); } } return 0; }