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            posts - 20,  comments - 6,  trackbacks - 0
              1#include<stdio.h>
              2#include<string.h>
              3int main()
              4{
              5    char nek[1000],bf[1000];
              6    int i,j,l,k;
              7    int as,bs,c1,c2;
              8    int ap,bp;
              9    int max;
             10    int t;
             11    
             12    while(scanf("%d",&t)!=EOF)
             13    {
             14        while(t--)
             15        {
             16            max=0;
             17            as=0;
             18            bs=1;
             19            scanf("%s",nek);
             20            strcpy(bf,nek);
             21            l=strlen(nek);
             22            while(bs<l)
             23            {
             24                ap=bp=0;
             25                c1=0;
             26                c2=0;
             27                for(i=as;;i--)
             28                {
             29                    if(i<0)
             30                    {
             31                        i=l-1;
             32                    }

             33                        if(nek[i]=='a')
             34                        break;
             35                    if(nek[i]=='w')
             36                        {
             37                            nek[i]='a';
             38                            c1++;
             39                        }

             40                        if(nek[i]=='p')
             41                        {
             42                            if(ap==0||ap==1)
             43                            {
             44                                nek[i]='a';
             45                            ap=1;
             46                            c1++;
             47                            }
            else
             48                                break;
             49                                
             50                        }

             51                        if(nek[i]=='r')
             52                        {
             53                            if(ap==0||ap==2)
             54                            {
             55                                nek[i]='a';
             56                            ap=2;
             57                            c1++;
             58                            }
            else
             59                            break;
             60                        }

             61                    
             62                            
             63                }

             64                for(j=bs;;j++)
             65                {
             66                    if(j==l)
             67                    {
             68                        j=0;
             69                    }

             70                        if(nek[j]=='a')
             71                        break;
             72                        if(nek[j]=='w')
             73                        {
             74                            nek[j]=='a';
             75                            c2++;
             76                        }

             77                        if(nek[j]=='p')
             78                        {
             79                            if(bp==0||bp==1)
             80                            {nek[j]=='a';
             81                            bp=1;
             82                            c2++;
             83                            }

             84                            else
             85                            break;                            
             86                        }

             87                        if(nek[j]=='r')
             88                        {
             89                            if(bp==0||bp==2)
             90                            {nek[j]=='a';
             91                            bp=2;
             92                            c2++;
             93                            }

             94                            else
             95                            break;
             96                        }

             97                    
             98    
             99                }

            100                if(max<c1+c2)
            101                {
            102                    max=c1+c2;
            103                }

            104                strcpy(nek,bf);
            105                c1=c2=0;
            106                ap=bp=0;
            107                as++;
            108                bs++;
            109            }

            110                c1=c2=0;
            111                ap=bp=0;
            112                as=bs=1;
            113                for(i=l-1,j=0;i>j;i--,j++)
            114                {
            115                    if(nek[i]=='w'&&as)
            116                    {
            117                        c1++;
            118                    }

            119                    if(nek[j]=='w'&&bs)
            120                    {
            121                        c2++;
            122                    }

            123                    if(nek[i]=='p'&&as)
            124                    {
            125                        if(ap==0||ap==1)
            126                        {
            127                        ap=1;
            128                        c1++;
            129                        }
            else
            130                        {as=0;
            131                        }

            132                    }

            133                    if(nek[j]=='p'&&bs)
            134                    {
            135                        if(bp==0||bp==1)
            136                        {
            137                        bp=1;
            138                        c2++;
            139                        }
            else
            140                        {
            141                            bs=0;
            142                        }

            143                    }

            144                    if(nek[i]=='r'&&as)
            145                    {
            146                        if(ap==0||ap==2)
            147                        {
            148                        ap=2;
            149                        c1++;
            150                        }
            else
            151                        as=0;
            152                    }

            153                    if(nek[j]=='r'&&bs)
            154                    {
            155                        if(bp==0||bp==2)
            156                        {
            157                        bp=2;
            158                        c2++;
            159                        }
            else
            160                        {
            161                            bs=0;
            162                        }

            163                    }

            164                }

            165                if(max<c1+c2)
            166                {
            167                    max=c1+c2;
            168                }

            169                printf("%d\n",max);
            170        }

            171    }

            172}

            173                
            174            
            posted on 2009-01-27 14:03 混沌的云 閱讀(124) 評論(0)  編輯 收藏 引用
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