Description
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.
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
28
思路：假设n为数字个数。如果n为奇数，就选n/2个最大的数从大到小排，剩下的从小到大排（不能有前导零）；如果n为偶数，首先选相邻两位的差最小的两个，然后大的那个后面就接小的（从小到大排），小的那个就接大的（从大到小排）。
#include <stdio.h>
bool vis[10];
int num[10];
int main()
{
int t,n,i,j,cnt,num1,num2,minsub,min;
char s[20];
scanf("%d\n",&t);
while(t--)
{
gets(s);
n=0;
for(i=0;s[i]!=0;i++) if(s[i]==‘ ‘) n++;
n++;
for(i=0;i<n;i++) sscanf(&s[i*2],"%d",&num[i]);
num1=num2=0;
if(n%2)//如果是奇数个
{
cnt=n/2;
for(i=n-1;i>=n-cnt;i--) num2=num2*10+num[i];
cnt++;
if(!num[0])//有0
{
num1=num1*10+num[1];
num1*=10;
cnt-=2;
for(i=0;i<cnt;i++) num1=num1*10+num[2+i];
}
else
{
for(i=0;i<cnt;i++) num1=num1*10+num[i];
}
printf("%d\n",num1-num2);
}
else//偶数个
{
if(n==2)//只有2个
{
printf("%d\n",num[1]-num[0]);
}
else//4个以上
{
minsub=min=999999999;
for(i=0;i<n-1;i++) if(num[i+1]-num[i]<minsub && num[i]) minsub=num[i+1]-num[i];
for(i=0;i<n-1;i++) if(num[i+1]-num[i]==minsub && num[i])//枚举相邻两位数的差最小的所有情况
{
for(j=0;j<10;j++) vis[j]=0;
num1=num[i+1];
num2=num[i];
vis[num[i+1]]=vis[num[i]]=1;
cnt=n/2-1;
for(j=0;cnt>0;j++)
{
if(!vis[num[j]])
{
vis[num[j]]=1;
cnt--;
num1=num1*10+num[j];
}
}
cnt=n/2-1;
for(j=n-1;cnt>0;j--)
{
if(!vis[num[j]])
{
vis[num[j]]=1;
cnt--;
num2=num2*10+num[j];
}
}
if(num1-num2<min) min=num1-num2;
}
printf("%d\n",min);
}
}
}
}