USACO "bead" problem

rapunzel (11)
This is "USACO" bead problem, why doesn't it print the number?
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#include<iostream>
#include<fstream>
using namespace std;
int main()
{
    int n,i,out=0,w=0,pre,k;
    cin>>n;
    char a[n];
    for(int q=0;q<n;q++)
    {
            cin>>a[q];
    }
    while(w<n)
    {
              pre=0;         
              for(int r=w;r>=0;r++)
              {
                      if(r==n)r=0;
                      if(a[w]='r')
                      {
                                  if(a[r]=='b')break;
                      }
                      if(a[w]=='b')
                      {
                                   if(a[r]=='r')break;
                      }
                      pre++;
                      k=r+1;
              }    
              for(int e=k;e>=0;e++)
              {
                      if(e==n)e=0;
                      if(a[w]='r')
                      {
                                  if(a[e]=='r')break;
                      }
                      if(a[w]=='b')
                      {
                                   if(a[e]=='b')break;
                      }
                      pre++;       
              } 
              if(pre>out)out=pre;
              w++;  
    }
    cout<<out<<endl;
    cin>>i;
    return 0;
}

Last edited on
rapunzel (11)
PLEASE HELP MEEEEE . . .
It's EMERGENCY :(
AbstractionAnon (6954)
If you never execute the break statements, you're never going to exit the loop since r>=0 will always be true.
rapunzel (11)
Thank you but I understood my answer .

If you want to know the correct answer of it (that accepted from USACO)you can see it here :
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/*
ID:
PROG: beads
LANG: C++
*/
#include<iostream>
#include<fstream>
using namespace std;
int main()
{
     ofstream fout ("beads.out");
    ifstream fin ("beads.in");
    int n,i,out=0,w=0,pre,k;
    fin>>n;
    char a[n],num1,num2;
    for(int q=0;q<n;q++)
    {
            fin>>a[q];
    }
    for(int l=0;l<n;l++)
    {
            w=l;
            pre=0;
            num1=a[w];         
            for(int r=w;r>=0;r++)
            {
                    if(r==n)r=0;
                    num2=a[r];
                    if(num1=='r')
                    {
                                if(num2=='b') break;
                    }
                    if(num1=='b')
                    {
                                 if(num2=='r') break;
                    }
                    pre++;
                    if(num1=='w')num1=num2;
                    if(pre==n)
                    {
                              break;    
                    }
            }
            k=pre+w;
           if(k>=n)k=k%n; 
            num1=a[k]; 
            if(n>pre)
            {
                      for(int e=k;e>=0;e++)
                      {
                              
                              num2=a[e];
                              if(e==n-1)e=-1;
                              if(num1=='r')
                              {
                                           if(num2=='b')break;
                              }
                              if(num1=='b')
                              {
                                           if(num2=='r')break;
                              }       
                              if(num1=='w')num1=num2;
                              pre++;      
                      }
            }
            if(pre==n)
            {
                      out=pre;
                      break;
            }
            else if(pre>out)
            {
                       out=pre;
            } 
    }
    fout<<out<<endl;
    return 0;
}


if you used it please tell me ,I will be happy if I find out that my program would be helpful:)
Last edited on
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