1、　　编制程序，将输入的一行字符加密和解密。加密时，每个字符依次反复加上“4962873”中的数字，如果范围超过ＡＳＣＩＩ码的032（空格）~122（‘z’），则进行模运算。解密与加密的顺序相反。编制加密和解密函数，打印各个过程的结果。（输出结果见课本183页8.4）-1, the preparation process will be a row of characters input encryption and decryption. Encryption, each character in turn repeated plus " 4962873" , if the range of more than 032 of the ASCII code (spaces) to 122 (' z' ), modular arithmetic. Decryption and encryption in reverse order. Preparation of encryption and decryption functions, print the results of the various processes. (See textbooks 183 8.4 output)