http://pat.zju.edu.cn/contests/pat-practise/1005

 

 

1005. Spell It Right (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

12345

Sample Output:

one five

反转字符串都要自己写！我了个去去去

[cpp]  view plain copy

1. #include<iostream>  
2. #include<cstdio>  
3. #include<memory.h>  
4. #include<algorithm>  
5. #include<cstring>  
6. #include<queue>  
7. #include<cmath>  
8. #include<cstdlib>  
9. #include<string>  
10. using namespace std;  
11. #define MAX 10000000000  
12. long long a[25];  
13. char s[102];  
14. char v[10][8] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};  
15. string ltoa(long long x){  
16.     string res;  
17.     while(x){  
18.         res += x%10 + '0';  
19.         x /= 10;  
20.     }  
21.     int len = res.length();  
22.     for(int i=0;i<len/2;++i){  
23.         char c = res[i];  
24.         res[i] = res[len-1-i];  
25.         res[len-1-i] = c;  
26.     }  
27.     return res;  
28. }  
29. int main(){  
30.   
31.     //freopen("in.txt", "r", stdin);  
32.     int j;  
33.     while(cin>>s){  
34.         memset(a, 0, sizeof(a));  
35.         int len = strlen(s);  
36.         for(int i=0;i<len;++i){  
37.             a[0] += s[i] - '0';  
38.             j = 0;  
39.             while(a[j]>MAX){  
40.                 a[j+1] += 1;  
41.                 a[j] = a[j]%MAX;  
42.                 j++;  
43.             }  
44.         }  
45.         j = 15;  
46.         while(j>=0 && a[j]==0){  
47.             j--;  
48.         }  
49.         if(j==-1){  
50.             printf("zero\n");  
51.             continue;  
52.         }  
53.         bool first = true;  
54.         while(j>=0){  
55.             string str = ltoa(a[j]);  
56.           
57.             for(int i=0;i<str.length();++i){  
58.                 if(first){  
59.                     first = false;  
60.                 }else{  
61.                     printf(" ");  
62.                 }  
63.                 printf("%s", v[str[i]-'0']);  
64.             }  
65.               
66.             j--;  
67.         }  
68.         printf("\n");  
69.     }  
70.     //fclose(stdin);  
71.     return 0;  
72. }