hdu Numerically Speaking

Numerically Speaking

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 163    Accepted Submission(s): 97

Problem Description

A developer of crossword puzzles (and other similar word games) has decided to develop a mapping between every possible word with from one to twenty characters and unique integers. The mapping is very simple, with the ordering being done first by the length of the word, and then alphabetically. Part of the list is shown below. a 1 b 2 ... z 26 aa 27 ab 28 ... snowfall 157,118,051,752 ...
Your job in this problem is to develop a program which can translate, bidirectionally, between the unique word numbers and the corresponding words.

 

Input

Input to the program is a list of words and numbers, one per line starting in column one, followed by a line containing a single asterisk in column one. A number will consist only of decimal digits (0 through 9) followed immediately by the end of line (that is, there will be no commas in input numbers). A word will consist of between one and twenty lowercase alphabetic characters (a through z).

 

Output

The output is to contain a single line for each word or number in the input data. This line is to contain the word starting in column one, followed by an appropriate number of blanks, and the corresponding word number starting in column 23. Word numbers that have more than three digits must be separated by commas at thousands, millions, and so forth.

 

Sample Input

29697684282993 transcendental 28011622636823854456520 computationally zzzzzzzzzzzzzzzzzzzz *

 

Sample Output

elementary 29,697,684,282,993 transcendental 51,346,529,199,396,181,750 prestidigitation 28,011,622,636,823,854,456,520 computationally 232,049,592,627,851,629,097 zzzzzzzzzzzzzzzzzzzz 20,725,274,851,017,785,518,433,805,270

 

Source

North Central North America 1997

 

Recommend

Eddy

 

分析：时隔一年再做这题，因为未清空数组还是没有1Y。但比其当年的不敢下手，自觉还是懂了挺多，大数模板也比较熟练了。

关于这道题OJ的数据普遍很弱，网上很多AC代码在输入26时不是得到z而是a`，这种情况是由于直接理解为了将10进制转化为26进制。其实不然。因为对于任意n进制，每位上的数都是0~n-1，但这题是1~26。所以，博主做法是，对26取模后，若为0，则将此位定为26，并把除以26的商-1后再进行下一步。

#include<cstdio>
#include<cmath>
#include<cstring>
int a[40], b[40], ca[40];

void add(int s[], int n) {
    int t = 0, i;
    s[39] += n;
    for (i = 39; i >= 0; --i) {
        s[i] += t;
        t = s[i] / 10;
        s[i] %= 10;
    }
}

void mul(int s[], int n) {
    int t = 0, i;
    for (i = 39; i >= 0; --i) {
        s[i] = s[i] * n + t;
        t = s[i] / 10;
        s[i] %= 10;
    }
}

void sub(int s[], int n) {
    int t = 0, i;
    s[39] -= n;
    for (i = 39; i >= 0; --i) {
        if (s[i] >= 0) {
            break;
        } else {
            s[i + 1]--;
            s[i] += 10;
        }
    }
}

int div_mod(int s[], int n) {
    int t = 0, i;
    for (i = 0; i < 40; ++i) {
        s[i] += 10 * t;
        t = s[i] % n;
        s[i] /= n;
    }
    return t;
}

int is_zero(int s[]) {
    int i;
    for (i = 0; i < 40; ++i)
        if (s[i])
            return 0;
    return 1;
}
char ss[40];

int main() {
    int le, i, ans;
    while (scanf("%s", ss) != EOF) {
        if (ss[0] == '*')
            break;
        le = strlen(ss);
        memset(a, 0, sizeof (a));
        memset(b, 0, sizeof (b));
        memset(ca, 0, sizeof (ca));
        if (ss[0] <= '9' && ss[0] >= '0') {
            for (i = 39; le > 0; --i)
                ca[i] = a[i] = ss[--le] - '0';

            for (i = 39; i >= 0; --i) {
                ans = div_mod(a, 26);
                if (!ans) {
                    b[i] = 26;
                    sub(a, 1);
                } else
                    b[i] = ans;
                if (is_zero(a))
                    break;
            }
            for (i = 0; i < 40; ++i)
                a[i] = ca[i];
        } else if (ss[0] <= 'z' && ss[0] >= 'a') {
            for (i = 39; le > 0; --i)
                b[i] = ss[--le] - 'a' + 1;
            for (i = 0; i < 40; ++i) {
                mul(a, 26);
                add(a, b[i]);
            }
        }
        for (i = 0; i < 40; ++i)
            if (b[i])

                break;
        ans = i;
        for (; i < 40; ++i)
            printf("%c", b[i] + 'a' - 1);
        ans -= 18;
        while (ans--)
            printf(" ");
        for (i = 0; i < 40; ++i)
            if (a[i])
                break;
        for (; i < 40; ++i) {
            printf("%c", a[i] + '0');
            if (i < 39 && i % 3 == 0)
                printf(",");
        }
        printf("\n");
    }
    return 0;
}