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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> C++ MySql實現個人電話本

C++ MySql實現個人電話本

編輯:C++入門知識

C++ MySql實現個人電話本


 

Language: Smallest Difference Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5034   Accepted: 1382

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.

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

Source

Rocky Mountain 2005
題意:給2~10個不同的數字,要求把這些數字分成兩堆,每一堆任意組合成一個數(開頭不能為0),求兩個數差的最小值。

 

思路:數據很小,直接暴力枚舉所有情況,並且很容易想到,要使差最小則這兩個數的位數應該盡量相等;枚舉排列時用到了next_permutation函數。

代碼:

 

#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
#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
)
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())!='
')
            if (ch>='0'&&ch<='9')
                num[cnt++]=ch-'0';
        if (cnt==2){
            pf(%d
,abs(num[1]-num[0]));
            continue;
        }www.Bkjia.com
        int ans=INF;
        while (num[0]==0)
            next_permutation(num,num+cnt);
        do{
            int mid=(cnt+1)>>1;
            if (num[mid])
            {www.Bkjia.com
                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
,ans);
    }
    return 0;
}


 

 

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