《Cracking the Coding Interview》——第17章：普通题——题目7

// 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 \n");
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('\n');
}

return 0;
}