Today is also written by my boyfriend! ! !
7-11 Counting numeric characters and spaces (10 points)
This question requires writing a program to enter a line of characters, and count the number of numeric characters, spaces and other characters. It is recommended to use the switch statement to write.
Input format:
Input several characters in one line, and the last carriage return indicates the end of input, which is not counted.
Output format:
follow in one line
blank = the number of blanks, digit = the number of digit characters, other = the number
of other characters , output in the format. Note that there is a space on the left and right of the equal sign, and a space after the comma.
Input sample:
Here is a set of input. E.g:
Reold 12 or 45T
output example:
Here is the corresponding output. E.g:
blank = 3, digit = 4, other = 8
#include <iostream>
using namespace std;
#include<string>
void ccc(char abc)
{
int a=0,b=0,c=0;
while((abc=cin.get())!='\n')
{
if(abc>='0'&&abc<='9')
{
a++;
}
else if(abc==' ')
{
b++;
}
else
c++;
}
cout<<"blank = "<<b<<","<<' '<<"digit = "<<a<<","<<' '<<"other = "<<c <<endl;
}
int main ()
{
char abc;
ccc(abc);
return 0;
}
There is also a boyfriend who wrote along my line of thought~
#include <iostream>
using namespace std;
#include<string>
void ccc(char abc[],int n)
{
int a=0,b=0,c=0;
int i;
for (i=0;i<n;i++)
{
if(abc[i]>='0'&&abc[i]<='9')
{
a++;
}
else if(abc[i]==' ')
{
b++;
}
else
{
c++;
}
}
cout<<"blank = "<<b<<","<<" digit = "<<a<<","<<" other = "<<c<<endl;
}
int main()
{
char abc[1000];
char ch;
int i;
int n=0;
ch=getchar();
for(i=0;ch!='\n';i++)
{
abc[i]=ch;
n++;
ch=getchar();
}
ccc(abc,n);
return 0 ;
}
Array writing, the idea is almost the same~
Thanks again boyfriend! ! ! !