Smallest Difference
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6493 | Accepted: 1771 |
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.
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.
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
给你 《=10个数，从中可以任意取出几个数，余下的组成另一个数，试问这两个数的最小差是多少
有几个注意的地方：1.对于给定的序列，，要让其差最小，那么取出这的两个数位数则需要尽量向相同靠齐
2.对于其中的一个序列的模拟，，是可以通过DFS将取出相同数字的所有不同放法遍历到的，所以
不需要双重DFS
3。对于已经取出的数字，可以直接在循环时就求出其值，不需要再弄个单独的函数，对数组遍历进行求值
4.特别要注意的是，，在dfs中一定要用if else if 语句，，否则会在 st==n/2+1之后
也就是取出符合条件的数字之后，又一次进入循环内，，所以最后tle了好几次，，就少了个
else ,,,
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define inf 0x3f3f3f3f
using namespace std;
int a[12],b[12],s[12],minn,v[12],n;
int dfs(int st,int va) //3
{
if(a[1]==0&&st>=2)
return 0;
if(st==n/2+1)
{
int r=0;
for(int j=1;j<=n;j++)
if(!v[j])
b[++r]=s[j];
do{
if(r>=2&&b[1]==0)
continue;
int temp=0;
for(int i=1;i<=r;i++)
temp=temp*10+b[i];
if(abs(temp-va)<minn)
minn=abs(temp-va);
}while(next_permutation(b+1,b+r+1));
}
else //4!!!!!!!!!!!!!!!!!没有这个TL;E了好几次
for(int i=1;i<=n;i++)
if(!v[i])
{
v[i]=1;
a[st]=s[i];
dfs(st+1,va*10+s[i]);
v[i]=0;
} //2
return 0;
}
void init()
{
memset(s,0,sizeof(s));
memset(v,0,sizeof(v));
n=0;
char m[30];
gets(m);
for(int i=0;m[i]!='\0';i++)
if(m[i]>='0'&&m[i]<='9')
s[++n]=m[i]-'0';
}
int main()
{
int cas;
scanf("%d\n",&cas);
while(cas--)
{
init();
minn=inf;
dfs(1,0);
printf("%d\n",minn);
}
return 0;
}