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[];
char s[];

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

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int N,K;
        int s1=,s2=,s3=;
        scanf("%d%d",&N,&K);
        scanf("%s",s+);
        for(int i=;i<=N;i++)
        {
            a[i]=s[i]-'';
            if(a[i]%==) a[i]=,s1++;
            else if(a[i]%==) a[i]=,s2++;
            else s3++,a[i]=(a[i])?:;
        }
        int ans=(s1+s2*)%;
        int a3,E1,E2,f=;
        cal(a3,E1,E2,N);
        for(int C=;C<=s2&&C<=K;C++)  ///C->2; B->1; A->0;
        {
            int B=((ans-C*)%+)%;
            for(;B<=s1&&C+B<=K;B=B+)
            {
                int A=K-C-B;
                if(A<=s3)
                {
                    if(A>a3) f=;
                    else if(B<s1&&E1) f=;
                    else if(C<s2&&E2) f=;
                    if(f) break;
                }
            }
            if(f) break;
        }
        if((N==K+)&&s3) f=;
        if(f) puts("yes");
        else puts("no");
    }
    return ;
}