Gym

Using at most 7 matchsticks, you can draw any of the 10 digits as in the following picture:
The picture shows how many sticks you need to draw each of the digits.

Zaytoonah has a number that consists of N digits. She wants to move some sticks (zero or more) to maximize the number. Note that she doesn’t want to remove any of the sticks, she will only move them from one place to another within the N digits. She also doesn’t want to add new digits as N is her lucky number.

Can you help Zaytoonah maximize her number?
Input

The first line of input contains a single integer T, the number of test cases.

Each test case contains a single integer N (1 ≤ N ≤ 105), followed by a space, then N digits that represent the number Zaytoonah currently has.
Output

For each test case, print on a single line the maximum number Zaytoonah can get.
Example
Input

3
1 3
3 512
3 079

Output

5
977

997

思路：

在木棍数量确定的前提下，通过移动次数，来使得摆出的数量最大，我们能够得到十个数字每一个数字所示有的木棍的个数（它们在2-7之间），那么在确定一根的前提下，剩下的根数需要在 n-1 * 2   ～ n-1*7 之间，每一位都从大到小进行枚举，即可找到最大的情况

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[15]={6,2,5,5,4,5,6,3,7,6};char s[100100];int main (){    int t,n,m;    scanf("%d",&t);    while(t--)    {        scanf("%d %s",&n, s);        m=strlen(s);        int sum=0;        for(int i=0; i<m;i++)        {            sum+=a[s[i]-'0'];        }        memset(s,0,sizeof(s));        for(int i=1;i<=n;i++)        {            for(int j=9;j>=0;j--)            {                if(sum-a[j]>=2*(n-i) && sum-a[j]<=7*(n-i))                {                    s[i-1]=j+'0';                    sum-=a[j];                    break;                }            }        }        puts(s);    }    return 0;}