ACM刷題之codeforces————The Little Match Girl

The Little Match Girl

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

用dfs做，做的時候記得剪枝，下面是ac代碼

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;

int x[10]={
	6,2,5,5,4,5,6,3,7,6
};

int fx[10]={
	6,7,3,5,4,2
};

int mx[10]={
	9,1,5,5,4,5,9,7,8,9
};

int gx[10]={
	0,0,1,7,4,5,9,8,0
};

char c[111111];
int a[111111];
int n;
bool f;
int wx[10];

bool dfs(int step,int sum)
{
	//printf("%d",q);
	if(f)
		return f;
	if(step==n&&sum==0){
		for(int i=0;i<n;i++)
		{
			printf("%d",a[i]);
		}
		printf("\n");
		return true;
	}
	if(sum<0||n==step)
	{
		return false;
	}
	
	if(sum/(n-step)<2||(sum>7*(n-step))) return false;
	
	for(int i=0;i<6;i++)
	{
		a[step]=gx[fx[i]];
		//printf("%d  %d\n",a[step],step);
		if(!f)
			f=dfs(step+1,sum-fx[i]);
		if(f)
			return f;
	}
	
}

int main()
{
	//freopen("f:/input.txt", "r", stdin);
	
	int zu,i,j,k,m,len1,sum,w;
	scanf("%d",&zu);
	while(zu--)
	{
		CLR(wx,0);
		scanf("%d %s",&n,c);
		sum=0;
		for(i=0;i<n;i++)
		{
			w=x[c[i]-'0'];
			sum+=(w);
			//++wx[w];
		}
		
		if(n==1)
		{
			printf("%d\n",mx[c[0]-'0']);
			continue;
		}
		
		
		//printf("%d\n",sum);
		f=false;
		dfs(0,sum);
	}
}