From keyboard input their phone number, according to the mobile phone number after more than two to three o, according to their under-pressure sieve analyzer output group, grouping situation is: to pray for more than 1, 3 is the first group; For more than 3 o 2, that is, for the second group; For more than 3 o to 0, for the third group, namely
I was just an introduction to programming,, every brother please help to give directions, thank you

CodePudding user response:

The code is as follows:

 
# include & lt; Iostream> 
using namespace std; 

Int main () 
{
Char name [50]. 
int i=0; 
int num; 
cout <"Please enter the phone number:"; 
Cin> name; 
While (the name [I]) + + I; 
Num=name] [I - 1-48; 
Num +=(name [2] I - - 48) * 10; 
Cout<" Now is the first "; 
The switch (num % 3) 
{
Case 0: cout<" Three groups "& lt; Case 1: cout<" A set of "& lt; Case 2: cout<" Two groups "& lt; } 
return 0; 
} 

Idea is, enter a string representing a phone number, and then find the end of the string, after a while, I will represent the '\ 0' position, then after the two natural is and I - I - 1, 2, will they add to the num integer variables, and then to take over operation of num, use the switch to branch output can complete the questions

CodePudding user response:

 # include & lt; stdio.h> 

Int main () 
{
Int ch1=0, ch2=0, ch. 
int num; 

Ch=ch1=ch2=0; 
While ((ch=getchar ())! )='\ n' {
Ch2=ch1;//save the number the penultimate number 
Ch1=ch;//save rent a number after 
} 
Printf (" % % c, c \ n ", ch1 and ch2); 
Num=((ch2 - '0') * 10 + ch1 - '0') % 3. 
Printf (" now is the first group % d \ n ", num==0? 3: num); 
return 0; 
} 

Pure C language code, please test,