POJ 2718

来源：互联网发布：剑网3插件数据编辑：程序博客网时间：2024/05/01 02:43

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;    }}

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