PAT (Advanced Level 甲级) Practice 1005 Spell It Right (20 分) C++实现

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该程序读取一个不超过100位的非负整数，计算其所有数字的和，并将结果的每个数字转换为英文单词输出。不使用itoa函数，而是手动实现了数字到英文单词的映射。

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 (≤10100).

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:

结尾无空行

Sample Output:

one five

结尾无空行

Answer：

注：PAT不支持itoa，否则这里的int转string，可以直接用itoa(tot,num,10）int转char实现。

#include<bits/stdc++.h>
#include<cstdlib>
using namespace std;
string num;
int main(){
	string a;
	cin>>a;
	int tot=0;
	for(int i=0;i<a.length();i++){
		tot+=(a[i]-'0');
	}
	stringstream snum;
	snum<<tot;
	num=snum.str();
	//itoa(tot,num,10);
	for(int i=0;i<=0;i++){
		if(num[i]=='0')cout<<"zero";
		if(num[i]=='1')cout<<"one";
		if(num[i]=='2')cout<<"two";
		if(num[i]=='3')cout<<"three";
		if(num[i]=='4')cout<<"four";
		if(num[i]=='5')cout<<"five";
		if(num[i]=='6')cout<<"six";
		if(num[i]=='7')cout<<"seven";
		if(num[i]=='8')cout<<"eight";
		if(num[i]=='9')cout<<"nine";
	} 
	for(int i=1;i<num.length();i++){
		if(num[i]=='0')cout<<" zero";
		if(num[i]=='1')cout<<" one";
		if(num[i]=='2')cout<<" two";
		if(num[i]=='3')cout<<" three";
		if(num[i]=='4')cout<<" four";
		if(num[i]=='5')cout<<" five";
		if(num[i]=='6')cout<<" six";
		if(num[i]=='7')cout<<" seven";
		if(num[i]=='8')cout<<" eight";
		if(num[i]=='9')cout<<" nine";
	} 

}