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.
Each input file contains one test case. Each case occupies one line which contains an N (≤10^100).
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.
12345
one five
#include <iostream> #include <string> #include <stack> using namespace std; string str[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; int main() { string s; cin >> s; int sum,i,temp; for (i=0; i< s.length(); i++) sum += (s[i]-‘0‘);//字符串转数字 temp = sum; stack<int> st; while(temp){ st.push(temp%10); temp/=10; } if (sum ==0) cout << str[0]; else { while (st.size()){ cout << str[st.top()]; if (st.size() != 1) cout << " "; st.pop(); } } return 0; }
#include <iostream> using namespace std; string arr[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; int main(){ string s,str; cin >> s; int i,sum = 0; for (i=0;i<s.length();i++) sum+= s[i]-‘0‘;//输入的字符串转为数字求和 str = to_string(sum);//把和转为字符串 cout << arr[str[0] - ‘0‘];//先输出第一个，如果为 0则直接输出 zero for (i=1; i < str.length(); i++) cout << " " << arr[str[i] - ‘0‘];//字符串转为整数作为数组下标输出对应英文 return 0; }
原文：https://www.cnblogs.com/cgy-home/p/15111550.html