Algorithm

D. Polycarp and Div 3
time limit per test
3 seconds
memory limit per test
256 megabytes
Problem link- https://codeforces.com/contest/1005/problem/D
input
standard input
output
standard output

Polycarp likes numbers that are divisible by 3.

He has a huge number s�. Polycarp wants to cut from it the maximum number of numbers that are divisible by 33. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after m� such cuts, there will be m+1�+1 parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by 33.

For example, if the original number is s=3121�=3121, then Polycarp can cut it into three parts with two cuts: 3|1|213|1|21. As a result, he will get two numbers that are divisible by 33.

Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.

What is the maximum number of numbers divisible by 33 that Polycarp can obtain?

Input

The first line of the input contains a positive integer s�. The number of digits of the number s� is between 11 and 2⋅1052⋅105, inclusive. The first (leftmost) digit is not equal to 0.

Output

Print the maximum number of numbers divisible by 33 that Polycarp can get by making vertical cuts in the given number s�.

Examples
input
Copy
3121
output
Copy
2
input
Copy
6
output
Copy
1
input
Copy
1000000000000000000000000000000000
output
Copy
33
input
Copy
201920181
output
Copy
4
Note

In the first example, an example set of optimal cuts on the number is 3|1|21.

In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by 33.

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and 3333 digits 0. Each of the 3333 digits 0 forms a number that is divisible by 33.

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers 00, 99, 201201 and 8181 are divisible by 33.

 



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 2e5 + 1;
int dp[N][3][2];
string s;

int rec(int idx, int mod, bool tak) {
  if(idx == s.length())
    return mod == 0;

  int &ret = dp[idx][mod][tak];
  if(ret != -1)
    return ret;
  ret = 0;

  ret = max(ret, (idx != 0 && mod == 0) + rec(idx + 1, (s[idx] - '0') % 3, s[idx] != '0'));
  ret = max(ret, rec(idx + 1, (mod + (s[idx] - '0')) % 3, tak));

  return ret;
}

int main() {
  cin >> s;

  memset(dp, -1, sizeof dp);
  cout << rec(0, 0, 0) << endl;

  return 0;
}
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Input

x
–
+
cmd
3121

Output

x
–
+
cmd
2
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Demonstration

Codeforces Solution-D. Polycarp and Div 3-Solution in C, C++, Java, Python,Polycarp and Div 3,Codeforces Solution

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