Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.
Input Specification:
Each input file contains one test case. Each case occupies one line which contains an N (≤10
100
).
Output Specification:
For each test case, output in one line the digits of the sum in English words. There must be one space between two consecutive words, but no extra space at the end of a line.
Sample Input:
12345
Sample Output:
one five
The initial individual explanations idea is to set a string constant, using the summation values obtained for your if else the calculated number corresponding output:
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int sum = 0;
char str[101];
const char* a[]={"zero","one","two","three","four","five","six","seven","eight","nine"};
cin>>str;
int len=strlen(str);
for(int i=0;i<len;++i)
{
sum = sum+(str[i]-'0');
}
//cout<<sum<<endl;
if(sum>=0&&sum<10)
{
printf("%s",a[sum%10]);
}
else if(sum>10&&sum<100)
{
printf("%s %s",a[sum/10],a[sum%10]);
}
else if(sum>100)
{
printf("%s %s %s",a[sum/100],a[(sum/10)%10],a[sum%10]);
}
return 0;
}More preferably the method is:
#include <iostream>
using namespace std;
int main() {
string a;
cin >> a;
int sum = 0;
for (int i = 0; i < a.length(); i++)
sum += (a[i] - '0');
string s = to_string(sum);
string arr[10] = {"zero", "one", "two", "three", "four", "five", "six",
"seven", "eight", "nine"};
cout << arr[s[0] - '0'];
for (int i = 1; i < s.length(); i++)
cout << " " << arr[s[i] - '0'];
return 0;
}to_stringFunction into the digital value of the output string the perfect solution for the problem at various locations