题面：
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 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
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 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 3333digits 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.
题目描述：
给你一个字符串，你可以将这个字符串切成任意部分，问你最多可以使得多少个部分的数%3为0
题目分析：
对于这个题来说，因为我们发现后一个数的答案可以由前一个数转移过来，因此我们可以考虑用dp去解决问题。
首先建立dp[i]和f[s]分别表示前i个字符串中能获得的最多符合条件的数的个数，f[s]代表将每一位数相加%3的余数为s的数中能获得最多符合条件的数的个数。
因为我们很容易发现，当某一位的个数为0的时候，最优的方案为将0单独作为余数。此时有dp[i]=dp[i-1]+1。而当某一位的数不为0时，此时，dp[i]=max(dp[i-1]+f[x]+1)。
由此状态转移方程就可以写出来了。
代码：
#include <bits/stdc++.h>
#define maxn 200005
using namespace std;
int dp[maxn],f[maxn];
char str[maxn];
int main()
{
scanf("%s",str+1);
int len=strlen(str+1);
f[1]=f[2]=-0x3f3f3f3f;
for(int i=1,s=0;i<=len;i++){
s=(s+str[i]-'0')%3;
if(str[i]=='0') dp[i]=dp[i-1]+1;
else dp[i]=max(dp[i-1],f[s]+1);
f[s]=max(f[s],dp[i]);
}
cout<<dp[len]<<endl;
return 0;
}