模拟题2(打表)

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hdu2816;
http://acm.hdu.edu.cn/showproblem.php?pid=2816;
题目意思很简单;并且这个题目应该也是很简单的,然而我却打了很久;特么还失误了几大块;
看来对这方面不熟悉哈;要多锻炼锻炼哈;;;
没什么说的直接摆代码吧;;

#include<stdio.h>#include<stdlib.h>#include<math.h>#include<iostream>#include<queue>#include<algorithm>#include<stack>#include<set>#include<string.h>using namespace std;//容易题;但并没有快速ac;;//两个失误点;//1;在打表的时候下标是字母时忘记打''单引号了;//2;最后转向的时候首先没有分奇偶的char zm[10][5],ch2[505],c[100];void db(){    zm[2][1]='A',zm[2][2]='B';zm[2][3]='C';    zm[3][1]='D',zm[3][2]='E';zm[3][3]='F';    zm[4][1]='G',zm[4][2]='H';zm[4][3]='I';    zm[5][1]='J',zm[5][2]='K';zm[5][3]='L';    zm[6][1]='M',zm[6][2]='N';zm[6][3]='O';    zm[7][1]='P',zm[7][2]='Q';zm[7][3]='R';zm[7][4]='S';    zm[8][1]='T',zm[8][2]='U';zm[8][3]='V';    zm[9][1]='W',zm[9][2]='X';zm[9][3]='Y';zm[9][4]='Z';}int main(){    char ch[1005],ch3[505];    int len,l,i,n1,n2,j;    db();    c['Q']='A';c['W']='B';c['E']='C';c['R']='D';c['T']='E';c['Y']='F';c['U']='G';    c['I']='H';c['O']='I';c['P']='J';c['A']='K';c['S']='L';c['D']='M';c['F']='N';    c['G']='O';c['H']='P';c['J']='Q';c['K']='R';c['L']='S';c['Z']='T';c['X']='U';    c['C']='V';c['V']='W';c['B']='X';c['N']='Y';c['M']='Z';    while(~scanf("%s",ch))    {        len = strlen(ch);        l = 0;        for(i = 0; i < len; i++)        {            n1=ch[i]-'0';            i++;            n2=ch[i]-'0';            //printf("%d %d\n",n1,n2);            ch2[l]=zm[n1][n2];            l++;        }        //for(i = 0; i < l; i++)        //{           // printf("%c",ch2[i]);        //}        //printf("\n");        for(i = 0; i< l ; i++)        {            ch2[i] = c[ch2[i]];            //printf("%c",ch2[i]);        }        //printf("l=%d\n",l);        int ll=0;        if(l==1)//最后反向的错了;;;;;;        {            ch3[ll++]=ch2[0];        }        else if(l%2==0)        {            //printf("isdss\n");            for(i =0;i < l/2;i++)            {                ch3[ll++] = ch2[i];                ch3[ll++] = ch2[l/2+i];            }        }        else        {             //printf("isdss\n");            ch3[ll++]=ch2[0];            for(i = 1; i<= l/2;i++)            {                ch3[ll++]=ch2[l/2+i];                ch3[ll++]=ch2[i];            }        }        for(i = ll-1; i>=0;i--)            printf("%c",ch3[i]);        printf("\n");    }    return 0;}
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