[排序]PAT1038 Recover the Smallest Number

1038. Recover the Smallest Number (30)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

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;}