CF1005D Polycarp and Div 3 思维

Polycarp likes numbers that are divisible by 3.

He has a huge number 3.

For example, if the original number is 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 0.

Output

Print the maximum number of numbers divisible by s.

Examples

input

Copy

output

Copy

input

Copy

output

Copy

input

Copy

1000000000000000000000000000000000

output

Copy

input

Copy

201920181

output

Copy

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 3.

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

题意：给我们一个数，我们可以对这个数进行任意的划分，但是不能出现前缀零的无意义数，问我们最多可以划分出几个可以整除三的数？

分析：直接遍历，考虑这四种情况就可以

1.如果单独的数能整除三，那么这数肯定可以，结果直接加一

2.如果不能整除三，那么将得到一个余数，然后接下来我们会在这个数的基础上加上新的数，如果余数和这个数的余数相同，则中间肯定加了一个能整除三的数，结果加一

3.如果不能整除三的数加上一个数后能整除三，则结果加一

4.如果不能整除三的数加上一个数既不能整除三且余数和之前不相同，则再加一个数肯定会得到和前面两个数相同的余数或者整除三

根据以上分析我们每次加数时将余数打上标志，然后根据上诉分析判断结果是否能加一，记得结果加一后要把标志置为0

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int main() {
    string s;
    while( cin >> s ) {
        ll now = 0, cnt = 0, vis[3] = { 1, 0, 0 };
        for( ll i = 0; i < s.length(); i ++ ) {
            now += s[i]-'0';
            now %= 3;
            if( vis[now] ) {
                cnt ++;
                now = 0;
                vis[1] = vis[2] = 0;
            } else {
                vis[now] = 1;
            }
        }
        cout << cnt << endl;
    }
    return 0;
}