codeforces 1005 D Polycarp and Div 3

http://www.elijahqi.win/archives/3936
Polycarp likes numbers that are divisible by 3.

He has a huge number s

s
. Polycarp wants to cut from it the maximum number of numbers that are divisible by 3

3
. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after m

m
such cuts, there will be m+1

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

3
.

For example, if the original number is s=3121

s=3121
, then Polycarp can cut it into three parts with two cuts: 3|1|21

3|1|21
. As a result, he will get two numbers that are divisible by 3

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

3
that Polycarp can obtain?

Input
The first line of the input contains a positive integer s

s
. The number of digits of the number s

s
is between 1

1
and 2⋅105

2⋅105
, inclusive. The first (leftmost) digit is not equal to 0.

Output
Print the maximum number of numbers divisible by 3

3
that Polycarp can get by making vertical cuts in the given number s

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 3

3
.

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and 33

33
digits 0. Each of the 33

33
digits 0 forms a number that is divisible by 3

3
.

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

0
, 9

9
, 201

201
and 81

81
are divisible by 3

3
其實並沒有dp直接貪心即可

我們考慮分類討論即可 我在比賽中被hack了 因爲我沒有判斷最後一個是什麼

討論情況1、遇到0一定切刀

2、如果前後兩個非0且不一樣一定切刀

3、 如果前後兩個相同則一定需要切刀 因爲一定會存在滿足條件的情況

注意最後的邊界特判

具體討論細節看代碼即可

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+20;
char s[N];int a[N],n;
int main(){
    freopen("d.in","r",stdin);
    scanf("%s",s+1);n=strlen(s+1);
    for (int i=1;i<=n;++i) a[i]=s[i]-'0',a[i]%=3;
    int ans=0,last=0;
    for(int i=1;i<=n;++i){
        if (!a[i]) {++ans,last=0;continue;}
        if (!last) {last=a[i];continue;}
        if (last!=a[i]) {++ans,last=0;continue;}
        if (i+1>n) break;last=0;++i;++ans;
    }printf("%d\n",ans);
    return 0;
}