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Language: Default Smallest Difference
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 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 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. 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 Source |
题意：给2~10个不同的数字，要求把这些数字分成两堆，每一堆任意组合成一个数（开头不能为0），求两个数差的最小值。
思路：数据很小，直接暴力枚举所有情况，并且很容易想到，要使差最小则这两个数的位数应该尽量相等；枚举排列时用到了next_permutation函数。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
int num[12];
int main()
{
int i,j;
int cas,cnt;
char ch;
sf(cas);
getchar();
while (cas--)
{
cnt=0;
while ((ch=getchar())!='\n')
if (ch>='0'&&ch<='9')
num[cnt++]=ch-'0';
if (cnt==2){
pf("%d\n",abs(num[1]-num[0]));
continue;
}
int ans=INF;
while (num[0]==0)
next_permutation(num,num+cnt);
do{
int mid=(cnt+1)>>1;
if (num[mid])
{
int x=0,y=0;
FRL(i,0,mid)
x=x*10+num[i];
FRL(i,mid,cnt)
y=y*10+num[i];
ans=min(ans,abs(x-y));
}
}while (next_permutation(num,num+cnt));
pf("%d\n",ans);
}
return 0;
}