Polycarp and Div 3
Polycarp likes numbers that are divisible by 3.
He has a huge number ss. 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 mm such cuts, there will be m+1m+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=3121s=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 ss. The number of digits of the number ss 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 ss.
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 6forms 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.
思路：字数太多索性直接看样例猜题意，大致意思就是分割字符串是的最多的分开的块能够被3除尽并求除尽的块的数量
那肯定每个块内的元素数量越少越好呀，于是遇见0或3的倍数直接分成一个，其他的就是单块和sum和处理的问题了，除去直接分的我们发现还剩下的有1，2，4，5，7，8，而这些数至多三个组合在一起一定能被3整除，于是用sum存分块和，用cnt存单块个数，只需sum或cnt有一个被3整除那么一定就能分了
代码：
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = (int)2e5 + 10;
char str[maxn];
int main()
{
scanf("%s",str + 1);
int ans = 0,cnt = 0,sum = 0;
int len = strlen(str + 1);
for (int i = 1;i <= len;i ++)
{
sum += str[i] - '0';
cnt ++;
if ((str[i] - '0') % 3 == 0)
{
ans ++;
sum = 0;
cnt = 0;
}
else if (sum % 3 == 0 || cnt % 3 == 0)
{
ans ++;
sum = 0;
cnt = 0;
}
}
printf("%d\n",ans);
return 0;
}