HDU 6020---MG loves apple (enumeration)

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problem DescriptionMG is a rich boy. He hasNApples, each with a value of V (0<=V<=9 ).
A valid number does not contain a leading zero, and these apples has just made a validNDigit number.
MG have the right and take awayKApples in the sequence, he wonders if there exists a solution:after exactly taking awayKApples, the validN−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 do you have help settle the problem and calculate the answer? InputThe first line was an integerTWhich indicates the case number. (1<=T<= )
And as for each case, there is2Integern(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 ofNwould not exceed1000000。
Then there isn  integers < Span id= "mathjax-span-74" class= "math" > x  in the next line, the i-th integer means the i-th gold ' s value (< Span id= "mathjax-span-79" class= "mn" >0<= x<= 9 ). OutputAs for each case, you need to output a.
If the solution exists, print "yes", else print "no". (excluding quotation marks) Sample Input25 2112304 21000 Sample OutputYesno Test Instructions:

Ideas:

The code is as follows:

#include <iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespacestd;inta[100005];Chars[100005];voidCalint&AMP;A3,int&e1,int&e2,intN) {A3=0; e1=0; E2=0;  for(intI=1; i<=n;i++)    {        if(a[i]==3) Break; if(a[i]==0) a3++; }     for(intI=1; i<=n;i++)    {        if(a[i]==0) Break; if(a[i]==1) e1=1; if(a[i]==2) e2=1; }    return ;}intMain () {intT; CIN>>u;  while(t--)    {        intn,k; ints1=0, s2=0, s3=0; scanf ("%d%d",&n,&K); scanf ("%s", s+1);  for(intI=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++; ElseS3++,a[i]= (A[i])?3:0; }        intAns= (s1+s2*2)%3; inta3,e1,e2,f=0;        Cal (A3,e1,e2,n);  for(intC=0; c<=s2&&c<=k; C + +)///c->2; b->1; a->0;        {            intB= ((ans-c*2)%3+3)%3;  for(; b<=s1&&c+b<=k; b=b+3)            {                inta=k-c-B; if(a<=S3) {                    if(A&GT;A3) f=1; Else if(B&LT;S1&AMP;&AMP;E1) f=1; Else if(C&LT;S2&AMP;&AMP;E2) f=1; if(f) Break; }            }            if(f) Break; }        if((n==k+1) &AMP;&AMP;S3) f=1; if(f) Puts ("Yes"); ElsePuts"No"); }    return 0;}

HDU 6020---MG loves apple (enumeration)