大数删除k位是否能整除3，MG loves apple（HDU）

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MG loves apple

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 442    Accepted Submission(s): 76

Problem Description

MG is a rich boy. He has  n apples, each has a value of V( 0<=V<=9). 

A valid number does not contain a leading zero, and these apples have just made a valid  N digit number. 

MG has the right to take away  K apples in the sequence, he wonders if there exists a solution: After exactly taking away  K apples, the valid  N−K digit number of remaining apples mod  3 is zero. 

MG thought it very easy and he had himself disdained to take the job. As a bystander, could you please help settle the problem and calculate the answer?

Input

The first line is an integer  T which indicates the case number.（ 1<=T<=60)

And as for each case, there are  2 integer  N(1<=N<=100000), K(0<=K < N) in the first line which indicate apple-number, and the number of apple you should take away.

MG also promises the sum of  N will not exceed  1000000。

Then there are  N integers  X in the next line, the i-th integer means the i-th gold’s value( 0<=X<=9).

Output

As for each case, you need to output a single line.

If the solution exists, print”yes”,else print “no”.(Excluding quotation marks)

Sample Input

2 5 2 11230 4 2 1000

Sample Output

yes no

 

Source

BestCoder Round #93

题意：求去掉$K$位数字后，不含前导零，且数字和是否能被三整除。

 

我们设$S0$、$S1$、$S2$分别为原串上$mod3=0$、$1$、$2$数字的个数。 我们假定删除取模后为$0$、$1$、$2$的数字各$a$、$b$、$c$个，则显然有$0<=A<=S0,0<=B<=S1,0<=C<=S2$且$K=A+B+C$且$Sum$ 

$mod3=(A\ast 0+B\ast 1+C\ast 2)mod3=(S0\ast 0+S1\ast 1+S2\ast 2)mod3=bias$ 。 枚举 $C$ 的值，我们可得

mod 3=(A*0+B*1+C*2)mod 3=(S0*0+S1*1+S2*2)mod 3=bias$Bmod3=(bias-C\ast 2)mod3,A=K-B-C$。如果有若干组$A,B$不逾界，可知这些$(A,B,C)$是在模意义下合法的解，但不一定满足没有前导零。

所以，对于【大于$0$的数】我们贪心地从后往前删除，对于$0$我们贪心地从前往后删除。

需要统计出：$E3$=第一个【$mod3=0$且非$0$的数】前$0$的个数（如果$mod3=0$且非$0$的数不存在，那么$a3$就取所有零的个数），$E1$=【第一个$0$前是否存在$mod3=1$的数】，$E2$=【第一个$0$前是否存在$mod3=2$的数】。

则以下情况满足任一种都能保证无前导零：$A>=E3$。$B<$$S1$且$E1$。$C<$$S2$且$E2$。

还需要加一种情况 满足k==n-1 也是yes:

AC代码：

#include
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#include
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#include
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再推荐一道好题：https://oj.ejq.me/problem/24  （大数整除6的最大位数）

我的博客http://blog.csdn.net/xiangaccepted/article/details/69951928有这道题的解法（用的是dp）；