Codeforces 1005DPolycarp and Div 3

D. Polycarp and Div 3

time limit per test

3 seconds

memory limit per test

256 megabytes

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 $3$. 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 $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.

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

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 $3$ 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 $1$ and $2\cdot {10}^5$, inclusive. The first (leftmost) digit is not equal to 0.

Output

Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.

Examples

input

3121

output

2

input

6

output

1

input

1000000000000000000000000000000000

output

33

input

201920181

output

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 $3$.

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$

題意：給你一串由數字組成的字符串（最長2e5），要你把它分割成一些數字，使得在這些數字中能被3整除的個數最多。

解法：考慮將每個數字模3以後的結果。

如果對於當前數字能被3整除（0、3、6、9）模3結果爲0的數，則直接個數加1

如果是兩個數字那就有四種可能(1,1) (2, 2) (1, 2) (2, 1)其中後兩個組合之和爲3能被3整除

如果是三個數字對於之前兩個數字的組合只剩下(1, 1)和(2, 2) 那麼下個數不管是1還是2都可以組成一個3的整數倍

所以總結一下有三種情況

1.當前數字模3爲0

2.現有的數字之和模3爲0（當前正在處理的）

3.有三個數字了一定可以做到模3爲0

然後一個一個數字處理即可。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int a[N];
map<int,int> mp;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    string s;
    cin >> s;
    int len = s.length();
    int sum = 0, n = 0, cnt = 0;
    for(int i = 0; i < len; i++)
    {
        int temp = (s[i] - '0') % 3;
        sum += temp;//計算當前的模
        n++;//處理的數字+1
        if(n == 3 || sum % 3 == 0 || temp % 3 == 0)//如果已經有3個數字或者sum模3爲0或者當前數字是3的倍數
        {
            sum = n = 0;//n和sum都置0
            cnt++;//則個數加一
        }
    }
    cout << cnt << endl;
    return 0;
}