《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 }