The 5th Zhejiang Provincial Collegiate Programming Contest---ProblemG：Give Me the Number

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2971

题意：将输入的英文数字表达转化为阿拉伯数字。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 char aa[30][10]= {"zero", "one","two", "three", "four", "five", "six",  "seven", "eight", "nine",
 5                   "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen","nineteen",
 6                   "twenty","thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"
 7                  };
 8 int bb[28]= {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,30,40,50,60,70,80,90};
 9 char cc[300][20];
10 int main() {
11     int t;
12     char s[250];
13     scanf("%d",&t);
14     getchar();//先吃掉t后面的回车
15     while(t--) {
16         memset(s,0,sizeof(s));
17         memset(cc,0,sizeof(cc));
18         gets(s);
19         int num=0,k=0;
20         int len=strlen(s);
21         for(int i=0; i<len; i++) {
22             if(i>=1&&s[i]==' '&&s[i-1]!=' ') {
23                 num++;//在’ ‘前算上一个字符串
24                 k=0;
25                 continue;
26             }
27             if(s[i]!=' ')
28                 cc[num][k++]=s[i];//装入一个字符串
29         }
30         int sum1=0,sum2=0;
31         int j;
　　　　　　　　//有前至后读入字符串
32         for(int i=0; i<=num; i++) {
33             for(j=0; j<28; j++) {
34                 if(strcmp(cc[i],aa[j])==0) {
35                     sum1+=bb[j];
36                     break;
37                 }
38             }
39             if(j=28) {
40                 if(strcmp(cc[i],"and")==0)
41                     continue;
42                 if(strcmp(cc[i],"thousand")==0) {
43                     sum1*=1000;
44                     sum2+=sum1;
45                     sum1=0;
46                 } else if(strcmp(cc[i],"hundred")==0)
47                     sum1*=100;
48                 else if(strcmp(cc[i],"million")==0) {
49                     sum1*=1000000;
50                     sum2+=sum1;
51                     sum1=0;
52                 }
53             }
54         }
55         printf("%d\n",sum2+sum1);
56     }
57     return 0;
58 }//善于使用strcmp函数，其实想到就很简单了