《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：

// 17.7 Read an integer in English.
#include <map>
#include <string>
using namespace std;

map<int, string> m;

void init()
{
    m[0] = "zero";
    m[1] = "one";
    m[2] = "two";
    m[3] = "three";
    m[4] = "four";
    m[5] = "five";
    m[6] = "six";
    m[7] = "seven";
    m[8] = "eight";
    m[9] = "nine";
    m[10] = "ten";
    m[11] = "eleven";
    m[12] = "twelve";
    m[13] = "thirteen";
    m[14] = "fourteen";
    m[15] = "fifteen";
    m[16] = "sixteen";
    m[17] = "seventeen";
    m[18] = "eighteen";
    m[19] = "nineteen";
    m[20] = "twenty";
    m[30] = "thirty";
    m[40] = "forty";
    m[50] = "fifty";
    m[60] = "sixty";
    m[70] = "seventy";
    m[80] = "eighty";
    m[90] = "ninety";
    int i, j;
    for (i = 2; i <= 9; ++i) {
        for (j = 1; j <= 9; ++j) {
            m[i * 10 + j] = m[i * 10] + "-" + m[j];
        }
    }
}

void readNumber(int n)
{
    if (n == 0) {
        return;
    }
    // here n is limited between [0, 999];
    int a, b, c;
    
    a = n / 100;
    b = n % 100 / 10;
    c = n % 10 / 1;
    
    if (a > 0) {
        printf("%s hundred ", m[a].c_str());
        if (b != 0 || c != 0) {
            printf("and ");
        }
    }
    if (b * 10 + c > 0) {
        printf("%s ", m[b * 10 + c].c_str());
    }
}

int main()
{
    init();
    int n, n0;
    
    while (scanf("%d", &n) == 1) {
        if (n == 0) {
            printf("zero 
");
            continue;
        }
        if (n < 0) {
            printf("minus ");
            n = -n;
        }
        n0 = n;
        if (n >= 1000000000) {
            readNumber(n / 1000000000);
            printf("billion ");
            n = n % 1000000000;
        }
        if (n >= 1000000) {
            readNumber(n / 1000000);
            n = n % 1000000;
            printf("million ");
        }
        if (n >= 1000) {
            readNumber(n / 1000);
            n = n % 1000;
            printf("thousand ");
        }
        if (n0 >= 1000 && n / 100 == 0) {
            printf("and ");
        }
        readNumber(n);
        putchar('
');
    }
    
    return 0;
}