PAT 甲1038 Recover the Smallest Number (30)

最新推荐文章于 2021-08-14 15:27:30 发布

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本文介绍了一个算法问题：如何从给定的一系列数字片段中构造出最小的可能数值。通过比较不同数字片段组合的可能性，实现了一个排序算法来确定最优的数字排列顺序。

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Given a collection of number segments, you are supposed to recover thesmallest number from them. For example, given {32, 321, 3214, 0229, 87},we can recover many numbers such like 32-321-3214-0229-87 or0229-32-87-321-3214 with respect to different orders of combinations ofthese segments, and the smallest number is 0229-321-3214-32-87.

Input Specification:

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

Output Specification:

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

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#define MaxSize 10010

using namespace std;

bool cmp(string A,string  B)
{
    return A+B<B+A;
}

int main()
{
    string str[MaxSize];
    string answer;
    int i,n;
    scanf("%d",&n);
    for(i=0;i<n;i++)
        cin>>str[i];
    sort(str,str+n,cmp);
    for(i=0;i<n;i++)
        answer+=str[i];
        /*
    while(answer.size()!=0 &&answer[0]=='0')
    {
        answer.erase(answer.begin());

    }

   if(answer.size()!=0)
        cout<<answer;
    */
    i=0;
    while(i<answer.size())
    {
        if(answer[i]!='0')
            break;
        i++;
    }
    if(i<answer.size())
        printf("%s",answer.c_str()+i);     //此处使用answer+i报错，注意要使用指针，
   else
        printf("0");

    return 0;
}