POJ 2718

题意:
对于一个数列，我们将数字分成两堆，求可能的情况下，两堆数字组成的数的差的最小值，排除前导0
题解:
将给出的数列进行字典序排列，然后将每个部分的转化成数字做差比较，因为字典序，所以枚举了所有数字

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
#define pr(x) cout<<#x<<" "<<x;
#define pl(x) cout<<#x<<" "<<x<<endl;
#include<math.h>
#include<algorithm>
#include<set>
#include<map>
#include<bitset>
#include<string>
using namespace std;
int calnum(int str[],int eindex){
    int a= 0;
    for(int i = 0 ;i<eindex;i++){
        a = str[i]+a*10;
    }
    return a;

}
int main(){
    int t;
    cin>>t;
    cin.ignore();
    while(t--){
        string s;
        getline(cin,s);
        s.erase(remove(s.begin(),s.end(),' '),s.end());
        int length = s.size();
        int nums[length];
        for(int i = 0;i<length;i++){
            nums[i] = s[i]-'0';
        }
        int numa[length];
        int numb[length];
        int bendindex;
        int ans=0x7f7f7f7f;
        if(length%2==0){
            bendindex = length/2;
        }
        else{
            bendindex = length/2+1;
        }
        do{
            for(int i = 0;i<length/2;i++){
                numa[i] = nums[i];
            }
            for(int i = length/2,j = 0;i<length;i++,j++){
                numb[j] = nums[i];
            }

            if((numa[0]==0&&length/2>1)||
                (numb[0]==0&&bendindex>1)
                ){continue;}
            else{
                int a = calnum(numa,length/2);
                int b = calnum(numb,bendindex);
                ans = min(ans,abs(a-b));

            }

        }while(next_permutation(nums,nums+length));

        cout<<ans<<endl;
    }

}