HDU 6020---MG loves apple (enumeration), hdu6020 --- mg

HDU 6020---MG loves apple (enumeration), hdu6020 --- mg

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Problem DescriptionMG 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, cocould you please help settle the problem and calculate the answer? InputThe first line is an integer T which indicates the case number. (1 <= T <= 60)
And 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 shoshould 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 ). OutputAs for each case, you need to output a single line.
If the solution exists, print "yes", else print "no". (Excluding quotation marks) Sample Input25 2112304 21000 Sample OutputYesno:

 

Ideas:

 

The Code is as follows:

#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;}