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 87Sample Output:
22932132143287
题意：给出多个最长8位的数字，求他们组成的最小数字。
思路：这个想法比较简单了，假设两个数字字符串是str1，str2，直接排序排序规则是 str1+str2<str2+str1，可以得到最小的组合数字，注意要去掉前导0，如果全是0则只留下一个0。
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
string str[10005];
bool cmp(string str1,string str2)
{
return str1+str2<str2+str1;
}
int main()
{
int n,i,k;
cin>>n;
for(k=0;k<n;k++)
{
cin>>str[k];
}
sort(str,str+n,cmp);
int tag=0;
for(i=0;i<n;i++)
{
for(k=0;k<str[i].size();k++)
{
if(tag==0&&str[i][k]!=‘0‘)
{
cout<<str[i][k];
tag=1;
}
else if(tag==1) cout<<str[i][k];
}
if(tag==1) break;
}
if(i==n&&tag==0) cout<<"0";
i++;
for(i;i<n;i++)
cout<<str[i];
cout<<endl;
return 0;
}
[排序]PAT1038 Recover the Smallest Number,布布扣,bubuko.com
[排序]PAT1038 Recover the Smallest Number
原文：http://blog.csdn.net/zju_ziqin/article/details/19966909