首页 > 代码库 > 《Cracking the Coding Interview》——第17章:普通题——题目7
《Cracking the Coding Interview》——第17章:普通题——题目7
2014-04-28 23:28
题目:给定一个数字,用英语把它读出来。
解法:ZOJ上有相反的题目。如果我要用中文读书来呢?
代码:
1 // 17.7 Read an integer in English. 2 #include <map> 3 #include <string> 4 using namespace std; 5 6 map<int, string> m; 7 8 void init() 9 { 10 m[0] = "zero"; 11 m[1] = "one"; 12 m[2] = "two"; 13 m[3] = "three"; 14 m[4] = "four"; 15 m[5] = "five"; 16 m[6] = "six"; 17 m[7] = "seven"; 18 m[8] = "eight"; 19 m[9] = "nine"; 20 m[10] = "ten"; 21 m[11] = "eleven"; 22 m[12] = "twelve"; 23 m[13] = "thirteen"; 24 m[14] = "fourteen"; 25 m[15] = "fifteen"; 26 m[16] = "sixteen"; 27 m[17] = "seventeen"; 28 m[18] = "eighteen"; 29 m[19] = "nineteen"; 30 m[20] = "twenty"; 31 m[30] = "thirty"; 32 m[40] = "forty"; 33 m[50] = "fifty"; 34 m[60] = "sixty"; 35 m[70] = "seventy"; 36 m[80] = "eighty"; 37 m[90] = "ninety"; 38 int i, j; 39 for (i = 2; i <= 9; ++i) { 40 for (j = 1; j <= 9; ++j) { 41 m[i * 10 + j] = m[i * 10] + "-" + m[j]; 42 } 43 } 44 } 45 46 void readNumber(int n) 47 { 48 if (n == 0) { 49 return; 50 } 51 // here n is limited between [0, 999]; 52 int a, b, c; 53 54 a = n / 100; 55 b = n % 100 / 10; 56 c = n % 10 / 1; 57 58 if (a > 0) { 59 printf("%s hundred ", m[a].c_str()); 60 if (b != 0 || c != 0) { 61 printf("and "); 62 } 63 } 64 if (b * 10 + c > 0) { 65 printf("%s ", m[b * 10 + c].c_str()); 66 } 67 } 68 69 int main() 70 { 71 init(); 72 int n, n0; 73 74 while (scanf("%d", &n) == 1) { 75 if (n == 0) { 76 printf("zero \n"); 77 continue; 78 } 79 if (n < 0) { 80 printf("minus "); 81 n = -n; 82 } 83 n0 = n; 84 if (n >= 1000000000) { 85 readNumber(n / 1000000000); 86 printf("billion "); 87 n = n % 1000000000; 88 } 89 if (n >= 1000000) { 90 readNumber(n / 1000000); 91 n = n % 1000000; 92 printf("million "); 93 } 94 if (n >= 1000) { 95 readNumber(n / 1000); 96 n = n % 1000; 97 printf("thousand "); 98 } 99 if (n0 >= 1000 && n / 100 == 0) { 100 printf("and "); 101 } 102 readNumber(n); 103 putchar(‘\n‘); 104 } 105 106 return 0; 107 }
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