Polycarp and Div 3(思维+贪心)

Polycarp and Div 3

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⋅105

, 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.

题意：

将一串数切成若干份，问最多能得到多少能被3整除的数（不能有前导零）

分析：

首先一个数是3的倍数，那么各个位的和能被3整除（其实就是各个位模三的余数和能被三整除）

根据题意让我们得到尽可能多的三的倍数，所以可以贪心，让我们分的三的倍数尽可能的小

1）我们考虑单独一个数字：，如果是3的倍数（模3=0）则ans++

2）否则考虑两个数字，前两个数字对3取模可以得到以下结果[1,1][1,2][2,1][2,2]，很明显中间两项可 以被三整除，因为他们余数和恰好为3

3）如果两个数字的时候是[1,1][2,2]这两种情况，我们再考虑第三个数字，如果第三个数字直接可以被三整除那么我们舍弃前两个，只要第三个就可以了，是一样的。
如果第三个数字并不能被三整除，三个数分别对三取模我们得到一下情况[1,1,1],[2,2,2],[2,2,1],[1,1,2]注意这四种情况是从2）的[1,1][2,2]中来，因为另两种[1,2][2,1]构成的数可被三整除，不需要再构造三位数。
那么我们分析[1,1,1],[2,2,2],[2,2,1],[1,1,2]这四种情况，前两种余数加起来后都能被三整除，后两种我们只需要后两个也能被三整除，因此当我们判断到第三个数的时候，一定可以得到被三整除，这是再重新计数判断。

因此对于整个字符串来说我们只需要三个一循环，判断余数或余数和是否能被三整除或者长度到了三即可

code：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 2e5+5;
char str[N];
int main(){
    scanf("%s",str);
    int r = 0,sum = 0,ans = 0,n = 0;
    for(int i = 0; i < strlen(str); i++){
        r = (str[i] - '0') % 3;
        sum += r;
        n++;
        if(r == 0 || sum % 3 == 0 || n == 3){
            ans++;
            sum = n = 0;
        }
    }
    printf("%d\n",ans);
    return 0;
}