poj2718

源链接：http://poj.org/problem?id=2718
此题数据不大，最简单的做法就是直接用next_permutation来对原数组进行排列组合，我们知道，如果要使两个和最小，那么它们的长度肯定差不多，所以我们就对排列后的数组进行分割（注意原数组长度的奇偶性）。
这道题目有几个要注意的地方：
1.就是输入的格式问题，我们输入的时候每一个case,是以’\n’结尾，而最后一个case,我们是以’-1’（即EOF）结尾的，要注意判断。
2.因为当某个数字开头为0时我们判断该数字是非法的，而对于如原数组只有2个数字，且为0，x时，那么我们要特判，否则会将0这个单独的数字判为非法。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll __int64const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#pragma comment(linker,"/STACK:1024000000,1024000000")int n,m,k;#define M 110#define N 500010#define Mod 258280327#define p(x,y) make_pair(x,y)int a[10];int vis[10];int getIt(int s,int e){    int res=0;    if(a[s] == 0)        return -1;    for(int i=s;i<e;i++){        res = res*10+a[i];    }    return res;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    int t;    sf(t);    int flag=1;    while(t--){        if(flag) getchar();        char ch;        k=0;        while((ch=getchar())!='\n'){            if(ch == ' ')                continue;            if(ch == -1)                break;            a[k++] = ch-'0';        }        if(k == 2){     //特判            printf("%d\n",abs(a[0]-a[1]));            continue;        }        int ans = INF;        int x,y;        do{             x = getIt(0,k>>1);     //偶数长度数组             y = getIt(k>>1,k);             if(x!=-1&&y!=-1) ans = Min(ans,abs(x-y));             x = getIt(0,k>>1+1);   //奇数长度数组             y = getIt(k>>1+1,k);            if(x!=-1&&y!=-1) ans = Min(ans,abs(x-y));        }while(next_permutation(a,a+k));        flag=0;        printf("%d\n",ans);    }    return 0;}