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
题目没有粘全。。。。。
关键是有技巧。。
就是所有的数在经过题目中的条件下，一定的循环后一定会变成0，这个时候单独处理就不会超时。。。
#include <iostream>
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int main()
{
int T;
int x,k;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&x,&k);
if(k==0)
printf("%d\n",x);
else
{
int ans;
while(k--)
{
ans = 0;
if(x==0)
{
if(k%2==0)
ans = 1;
else
ans = 0;
break;
}
while(x)
{
int j = x%10;
if(j==0||j==4||j==6||j==9)
ans+=1;
else if(j==8)
ans+=2;
else ans+=0;
x/=10;
}
x = ans;
}
printf("%d\n",ans);
}
}
return 0;
}