woj~24. Divide by Six（dp）

Input file: standard inputOutput file: standard output Time limit: 1 secondMemory limit: 512 mebibytes

A positive integer number n is written on a blackboard. It consists of not more than 10^510​5​​ digits. You have to transform it into a mogicalnumber by erasing some of the digits, and you want to erase as few digits as possible.

The number is lucky if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 6. For example, 0, 66,66666 are lucky numbers, and 00, 25, 77 are not.

Write a program which for the given $n$ will find a mogical number such that $n$ can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number $n$.

Print the length of your answer after the erasing.

If it's impossible to obtain a lucky number, print -1s.

Input

The first line of input contains $n$ -- a positive integer ( 1\le n \le 10^{100000}1≤n≤10​100000​​ ).

Output

Print one number — the number of your lucky number obtained by erasing as few as possible digits. If there is no answer, print -1s.

Example

Input 1

0010456

Output 1

4

Input 2

11

Output 2

-1s

考虑两种情况，即这点取或不取，保存最优状态

dp[i][j]=x,i代表当前数字的位置，j代表取余6的数（把整个字符串当成一个数，j*10+t进行分段取余），而x代表的是dp[i][j]对应的最大长度。 
给dp赋初值的时候，是-100005，这个是防止所有都除去的情况，另外就是状态转移方程有两个，表示当前位置取或不取。 

#include <iostream>#include<stdio.h>#include<string>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stdlib.h>#include<stdio.h>#include<map>#include<vector>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define ll long long#define inf 100005;using namespace std;char s[100005];int dp[100005][7];int main(){    scanf("%s",s+1);    int len=strlen(s+1);    for(int i=0;i<=len;i++)    {        for(int j=0;j<6;j++)        {            dp[i][j]=-inf;        }    }    int ans=-inf;    for(int i=1;i<=len;i++)if(s[i]=='0')ans=1;    for(int i=1;i<=len;i++)    {        int t=s[i]-'0';        if(t!=0)        dp[i][t%6]=1;        //不取        for(int j=0;j<=5;j++)            dp[i][j]=max(dp[i-1][j],dp[i][j]);        //取        for(int j=0;j<=5;j++)            dp[i][(j*10+t)%6]=max(dp[i-1][j]+1,dp[i][(j*10+t)%6]);        ans=max(ans,dp[i][0]);    }    if(ans>0)        printf("%d\n",ans);    else        printf("-1s\n");}