HDU 6020---MG loves apple（枚举）

题目链接

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

题意：

思路：

代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
int  a[100005];
char s[100005];

void cal(int &a3, int &E1,int &E2,int N)
{
    a3=0; E1=0; E2=0;
    for(int i=1;i<=N;i++)
    {
        if(a[i]==3) break;
        if(a[i]==0) a3++;
    }
    for(int i=1;i<=N;i++)
    {
        if(a[i]==0) break;
        if(a[i]==1) E1=1;
        if(a[i]==2) E2=1;
    }
    return ;
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int N,K;
        int s1=0,s2=0,s3=0;
        scanf("%d%d",&N,&K);
        scanf("%s",s+1);
        for(int i=1;i<=N;i++)
        {
            a[i]=s[i]-‘0‘;
            if(a[i]%3==1) a[i]=1,s1++;
            else if(a[i]%3==2) a[i]=2,s2++;
            else s3++,a[i]=(a[i])?3:0;
        }
        int ans=(s1+s2*2)%3;
        int a3,E1,E2,f=0;
        cal(a3,E1,E2,N);
        for(int C=0;C<=s2&&C<=K;C++)  ///C->2; B->1; A->0;
        {
            int B=((ans-C*2)%3+3)%3;
            for(;B<=s1&&C+B<=K;B=B+3)
            {
                int A=K-C-B;
                if(A<=s3)
                {
                    if(A>a3) f=1;
                    else if(B<s1&&E1) f=1;
                    else if(C<s2&&E2) f=1;
                    if(f) break;
                }
            }
            if(f) break;
        }
        if((N==K+1)&&s3) f=1;
        if(f) puts("yes");
        else puts("no");
    }
    return 0;
}