C语言、学习、考试大全3

1IN.DAT =>?@A00B'(CDE\rXX ;@P`pCHA0,?p\nG:IJKL-CDE M"*?NCDMASCIIO\rPQ4 L*R"*STNCDMASCIIOU"*?NVMCD"*STNCDMASCII\rOPQ4 L*R"*SWNCDMASCIIOU"*STNVMCDXYZ[?\]\r^STNCDS?NCDMASCIIOR_"*?NCDMASCUOUS`NVM\rCDUMVCDab'cd_CDEefMLgh0"*i]^MCDE&?jV'\r(CDEx x "*+,pwriteDat0-$%xx./01OUT10.DAT 0_k\rl01'cMmnG\roMpqrs80NCDt5uDvwxm0\rvoid CharConvA(void) /*5678*/\r{int I,j,strl;\rchar ch;\rfor(I=0;I<MAXLINE;I++)\r{strl=strlen(xx[I]);\rch=xx[I][strl-l];\rfor(j=strl-l yj>O;j-)\r{xx[I][j]R=4;\rxx[I]U]+=xx[I][j-l]y.\rxx[I][01+=ch;\rvoid CharConvA(void)\r{ int i,jy\rchar p,c;\rfor(i=0;i<MAXLINE;I++)\r{ p=xx[i] [strlen(xx[i])-1 ];\rc=xx[i][0];\rfor(j=strlen(xx[i])-1 yj>O;j-)\rxx[i][j]=(xx[i]fj]Q4)+xx[i][j-l];\rxx[i][0]=p+c; }