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KK位数字后，不含前导零，且数字和是否能被三整除。

 
我们设S0S0、S1S1、S2S2分别为原串上mod
3=0mod3=0、11、22数字的个数。
我们假定删除取模后为00、11、22的数字各Aa、Bb、Cc个，则显然有0<=A<=S0,0<=B<=S1,0<=C<=S20<=A<=S0,0<=B<=S1,0<=C<=S2且K=A+B+CK=A+B+C且Sum 
mod3=(A∗0+B∗1+C∗2)mod3=(S0∗0+S1∗1+S2∗2)mod3=bias。
枚举CC的值，我们可得
mod
3=(A*0+B*1+C*2)mod 3=(S0*0+S1*1+S2*2)mod 3=biasB
mod 3=(bias-C*2)mod 3,A=K-B-CBmod3=(bias−C∗2)mod3,A=K−B−C。如果有若干组A,BA,B不逾界，可知这些(A,B,C)(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00的数】我们贪心地从后往前删除，对于00我们贪心地从前往后删除。
需要统计出：a3E3=第一个【mod
3=0mod3=0且非00的数】前00的个数（如果mod
3=0mod3=0且非00的数不存在，那么a3a3就取所有零的个数），E1E1=【第一个00前是否存在mod
3=1mod3=1的数】，E2E2=【第一个00前是否存在mod
3=2mod3=2的数】。
则以下情况满足任一种都能保证无前导零：A>=a3A>=E3。B<B<S1S1且E1E1。C<C<S2S2且E2E2。

还需要加一种情况 满足k==n-1 也是yes:
AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<string>
#define LL long long int
#define inf 0x3f3f3f3f
#define N 100010
using namespace std;
char aa
;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,k,s0=0,s1=0,s2=0,E1=0,E2=0,E3=0,flagE3=1,flagE1=1,flagE2=1;
scanf("%d%d",&n,&k);
scanf("%s",aa);
int sum=0;
for(int i=0;i<n;i++)
{
int t=aa[i]-'0';
sum=(sum+t)%3;
if(flagE3&&t%3==0)
{
flagE1=0;
flagE2=0;
if(t==0)
E3++;
else
flagE3=0;
}
if(flagE2&&t%3==2)
E2=1;
if(flagE1&&t%3==1)
E1=1;
if(t%3==0)
s0++;
if(t%3==1)
s1++;
if(t%3==2)
s2++;
}
int ans=0;
for(int c=0;c<=s2;c++)
{
int minb=((sum-c*2)%3+3)%3;//(sum-c*2)%3可能为-1或-2或0，1，2加上3之后再取余3，余数一定为非负。
for(int b=minb;b<=s1;b+=3)
{
int a=k-b-c;
if(a>=0&&a<=s0)
{
if((a>=E3)||(b<s1&&E1)||(c<s2&&E2))
{
ans=1;
break;
}
if(k==n-1)
{
ans=1;
break;
}
}
}
}

if(ans)
printf("yes\n");
else
printf("no\n");
}
}

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）；