Function and Function（青岛icpc）

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本文介绍了一个关于数字封闭区域计数的递归函数，通过定义特定的递归规则，计算由数字组成的字符串中每个数字产生的封闭区域总数。文章提供了函数实现的示例代码，并解释了如何通过递归调用进行计算。

滴答滴答---题目链接

Function and Function

Time Limit: 1 Second Memory Limit: 65536 KB

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

Hint

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

Submit Status

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int a[10]= {1,0,0,0,1,0,1,0,2,1};
int ff(long long n)
{

    int sum=0;
    if(n==0)
        return 0;
    while(n>0)
    {
        sum+=a[n%10];
        n/=10;
    }
    return sum;
}
long long f(long long n,long long k,long long m)
{
    int a;
    if(k==0)
        return n;
    a=ff(n);
    if(n==0)
    {
        if(k%2)
            return 1;
        else return 0;
    }
    if(n==1)
    {
        if(k%2)
            return 0;
        else return 1;
    }
    return f(a,k-1,m);
}
int main()
{
    long long t,k,n;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld%lld",&n,&k);
        printf("%lld\n",f(n,k,k));
    }
    return 0;
}