Smallest Difference

最新推荐文章于 2020-05-06 10:37:47 发布

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探讨了如何从一组不重复的十进制数字中选取并形成两个整数，使这两个整数之间的绝对差值达到最小，涉及算法设计与实现，具体通过C++代码演示了解题过程。

Smallest Difference

Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can be written down in some order to form a second integer. Unless the resulting integer is 0, the integer may not start with the digit 0.

For example, if you are given the digits 0, 1, 2, 4, 6 and 7, you can write the pair of integers 10 and 2467. Of course, there are many ways to form such pairs of integers: 210 and 764, 204 and 176, etc. The absolute value of the difference between the integers in the last pair is 28, and it turns out that no other pair formed by the rules above can achieve a smaller difference.

Input

The first line of input contains the number of cases to follow. For each case, there is one line of input containing at least two but no more than 10 decimal digits. (The decimal digits are 0, 1, …, 9.) No digit appears more than once in one line of the input. The digits will appear in increasing order, separated by exactly one blank space.

Output

For each test case, write on a single line the smallest absolute difference of two integers that can be written from the given digits as described by the rules above.

Sample Input

1
0 1 2 4 6 7

Sample Output

C++编写：

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
using namespace std;

vector<int> num;

int solve()
{
    num.clear();
    string a;
    getline(cin,a);
    for(int i=0;i<a.size();i++)          //将每一行数据存储到动态数组num中
    {
        if(a[i]>='0' && a[i]<='9')
            num.push_back(a[i]-'0');
    }
    
    int ans=0x3fffffff;
    int m,n;
    do{
        m=0,n=0;
        for(int i=0;i<num.size()/2;i++)         //得到第一个理想位数的整数
        {
            m *= 10;
            m += num[i];
        }
        for(int j=num.size()/2;j<num.size();j++)         //得到第二个理想位数的整数
        {
            n *= 10;
            n += num[j];
        }
        if((num[0]==0 && m!=0) || (num[num.size()/2])==0 && n!=0)    continue;
        ans=min(ans,abs(n-m));
    }while(next_permutation(num.begin(),num.end()));      //将num中的数据全排列
    cout<<ans<<endl;
}
 
int main()
{
    ios::sync_with_stdio(false);
    int m;        //m代表案例数
    cin>>m;
    cin.ignore();
    while(m--)
        solve();
}