PAT A 1038. Recover the Smallest Number (30)

题目

Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given {32, 321, 3214, 0229, 87}, we can recover many numbers such like 32-321-3214-0229-87 or 0229-32-87-321-3214 with respect to different orders of combinations of these segments, and the smallest number is 0229-321-3214-32-87.

Input Specification:

Each input file contains one test case.  Each case gives a positive integer N (<=10000) followed by N number segments.  Each segment contains a non-negative integer of no more than 8 digits.  All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the smallest number in one line.  Do not output leading zeros.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

 

求组合后最小的数。

对数字串由“小”到“大”排序，排序的“大”、“小”依据为：

头对齐，由高位向低位比较，直到某一位不同；

如果某一个数字串结束，则截取较长的数字串的剩余部分和短的数字串比较。

排序后的数字串即为最小的数，注意全0的情况即可。

 

代码：

#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;int Form_put(string s);//格式化输出，去0bool cm(const string &s1,const string &s2);//比较，不是单纯的字符串比较int main(){int n,flag=0,i;//数量；是否已经输出过数据的标志，1为有输出过vector<string> num;char tempc[10];cin>>n;//输入数据for(i=0;i<n;i++){scanf("%s",tempc);num.push_back(tempc);}sort(num.begin(),num.end(),cm);//排序for(i=0;i<n;i++)//输出{if(flag==1)//有输出数据时直接输出printf("%s",num[i].c_str());else//没有输出过，考虑去0flag=Form_put(num[i]);}if(flag==0)//！！！数据全为0，最后需要输出0cout<<0;return 0;}bool cm(const string &s1,const string &s2)//比较{int i=0;string s3;while(i<s1.size()&&i<s2.size())//比较最长子串{if(s1[i]<s2[i])return true;else if(s1[i]>s2[i])return false;i++;}if(i<s1.size())//子串相同，s2短。比较s1后面部分与s2{s3=&s1[i];return cm(s3,s2);}else if(i<s2.size())//子串相同，s1短。比较s1与s2后面部分{s3=&s2[i];return cm(s1,s3);}else//长度相同，位置无所谓return false;}int Form_put(string s)//格式化输出，补0，返回是否有数据输出{int i=0;while(i<s.size()&&s[i]=='0')i++;if(i<s.size()){cout<<&s[i];return 1;}elsereturn 0;}