ZOJ 4070 Function and Function

这题卡输入输出，不能用 cin 和  cout。

Function and Function

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Time Limit: 1 Second      Memory Limit: 65536 KB

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If we define , do you know what function means?

Actually, calculates the total number of enclosed areas produced by each digit in . The following table shows the number of enclosed areas produced by each digit:

Enclosed Area	Digit	Enclosed Area	Digit
0	1	5	0
1	0	6	1
2	0	7	0
3	0	8	2
4	1	9	1

For example, , and .

We now define a recursive function by the following equations:

For example, , and .

Given two integers and , please calculate the value of .

Input

There are multiple test cases. The first line of the input contains an integer (about ), indicating the number of test cases. For each test case:

The first and only line contains two integers and ( ). Positive integers are given without leading zeros, and zero is given with exactly one '0'.

Output

For each test case output one line containing one integer, indicating the value of .

Sample Input

6
123456789 1
888888888 1
888888888 2
888888888 999999999
98640 12345
1000000000 0

Sample Output

5
18
2
0
0
1000000000

Hint

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Author: WENG, Caizhi
Source: The 2018 ACM-ICPC Asia Qingdao Regional Contest

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;

int main()
{
	LL a[] = {1,0,0,0,1,0,1,0,2,1};
	LL n,m,j,k,i,T;
	scanf("%lld",&T);
    for (j=0;j<T;j++)
	{
		scanf("%lld%lld",&n,&m);
		if (m==0)
		{
			printf("%lld
",n);
			continue;
		}
		LL sum=0,x=0,flag = -1;
		for (i=0;i<m;i++)
		{
			if (n == 0)
			{
				flag = 0;
				break;
			}

			if (n == 1)
			{
				flag = 1;
				break;
			}
			sum=0;
			while (n)
			{
				sum = sum + (a[n%10]);
				n/=10;
			}
			n = sum;
			//cout<<"n = "<<n<<endl;
		}

		if (flag == 1)
		{
			if ( ( (m-1) - i) % 2 == 0 )
			printf("0
");
			else
			printf("1
");
		}
		else if (flag == 0)
		{
			if ( ( (m-1) - i) % 2 == 0 )
			printf("1
");
			else
			printf("0
");
		}
		else
		printf("%lld
",sum);
	}

	return 0;
}