ACM&a+b
A + B
- 题目描述:
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读入两个小于100的正整数A和B,计算A+B.
需要注意的是:A和B的每一位数字由对应的英文单词给出.
- 输入:
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测试输入包含若干测试用例,每个测试用例占一行,格式为"A + B =",相邻两字符串有一个空格间隔.当A和B同时为0时输入结束,相应的结果不要输出.
- 输出:
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对每个测试用例输出1行,即A+B的值.
- 样例输入:
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one + two = three four + five six = zero seven + eight nine = zero + zero =
- 样例输出:
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3 90 96
http://ac.jobdu.com/problem.php?pid=1010main.cpp1 #include <iostream> 2 #include <cstring> 3 4 using namespace std; 5 6 /************************ 7 8 one 9 10 two three 11 12 four five 13 14 six seven 15 16 eight 17 18 nine 19 20 zero 21 22 ************************/ 23 24 int trans(char a[], int i); 25 int num_len(int i); 26 27 int main() 28 { 29 char a[50]; 30 int len,x,y,i,j; 31 while(cin.getline(a,50)) { 32 len = strlen(a); 33 x=0; 34 y=0; 35 for(i=0; i<len&&a[i]!='+'; ) { 36 if(a[i]!=' ') { 37 j=trans(a,i); 38 x=x*10+j; 39 i+=num_len(j); 40 } 41 else i++; 42 } 43 for(i++;i<len&&a[i]!='='; ) { 44 if(a[i]!=' ') { 45 j=trans(a,i); 46 y=y*10+j; 47 i+=num_len(j); 48 } 49 else i++; 50 } 51 if(x==0&&y==0) 52 break; 53 cout<<x+y<<endl; 54 } 55 return 0; 56 } 57 58 int trans(char a[], int i) { 59 switch(a[i]) { 60 case 'o': return 1; break; 61 case 't': return a[i+1]=='w' ? 2 : 3; break; 62 case 'f': return a[i+1]=='o' ? 4 : 5; break; 63 case 's': return a[i+1]=='i' ? 6 : 7; break; 64 case 'e': return 8; break; 65 case 'n': return 9; break; 66 case 'z': return 0; 67 break; 68 } 69 return -1; 70 } 71 72 int num_len(int i) { 73 switch(i) { 74 case 1: return 4; break; 75 case 2: return 4; break; 76 case 3: return 5; break; 77 case 4: return 5; break; 78 case 5: return 5; break; 79 case 6: return 4; break; 80 case 7: return 6; break; 81 case 8: return 6; break; 82 case 9: return 5; break; 83 case 0: return 5; break; 84 } 85 return 0; 86 }
posted on 2013-03-12 22:12 Zachary_wiz 阅读(178) 评论(0) 编辑 收藏 举报