标签：sim blank eof return har ros turn problem bsp
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 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).
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.
29697684282993
transcendental
28011622636823854456520
computationally
zzzzzzzzzzzzzzzzzzzz
*
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
思路：可以看成是十进制到二十六进制的转化，包含了大整数的加减乘除，全程用手绘算法。。。
代码：
#include <iostream> #include <cstring> #include <cmath> #include <cstdio> using namespace std; char a[1000]; int b[1000]; int num[1005]; char d[1005]; char z[1005]; void fun1() { memset(num, 0, sizeof(num)); int O; for(int i = strlen(a) - 1; i >= 0; i--) { int o = 1000; int len = 0; b[0] = a[i] - ‘a‘ + 1; int rem = 0; for(int k = 0; k <= strlen(a) - 1 - i; k++) { for(int j = 0; j <= len; j++) { rem += b[j]*26; if(k == 0)rem /= 26; b[j] = rem%10; rem = rem/10; } while(rem != 0) { b[++len] = rem%10; rem = rem/10; } } int l = 0; int sum = 0; while(l <= len) { sum += num[o] + b[l]; num[o] = sum%10; sum = sum/10; o--; l++; if(l == len+1 && sum != 0){num[o] = sum%10;o--;} } O = o; } for(int i = 0; i < strlen(a); i++) cout << a[i]; for(int i = strlen(a); i < 22; i++) cout << ‘ ‘; int ji = (1000 - (O))%3; int j = 0; for(int q = O+1; q <= 1000; q++) { j++; cout << num[q]; if(1000 - O > 3) { if(j%3 == 0 && q != 1000)cout << ‘,‘; if(j == ji){cout << ‘,‘;j = 0;ji = -1;} } } cout << endl; } void fun2() { memset(z, 0, sizeof(z)); memset(d, 0, sizeof(d)); int len = strlen(a); strcpy(z, a); int j = 0; int bu = 0; while(1) { int sum = 0; int s = 0; int haha = 0; for(int i = 0; i < len; i++) { sum += a[i] - 48; int temp = sum/26; if(haha == 0 && temp != 0)haha = 1; if(haha == 1)a[s++] = temp + 48; sum = sum%26; sum *= 10; } len = s; sum /= 10; if(sum <= 0){d[j] = 26 + ‘a‘ - 1; bu = -1;j++;} else {d[j++] = sum + ‘a‘ - 1 + bu;bu = 0;} if(len <= 0) break; } for(int i = j-1; i >= 0; i--) cout << d[i]; for(int i = 0; i < 22-j; i++) { cout << ‘ ‘; } int ji = 0; int ans = strlen(z)%3; for(int i = 0; i < strlen(z); i++) { ji++; cout << z[i]; if(strlen(z) > 3) { if(ji%3 == 0 && i != strlen(z) - 1)cout << ‘,‘; if(i+1 == ans && ans != 0){cout << ‘,‘;ji = 0;} } } cout << endl; } int main() { while(cin >> a) { if(a[0] == ‘*‘)break; if(a[0] >= ‘0‘ && a[0] <= ‘9‘)fun2(); else if(a[0] >= ‘a‘ && a[0] <= ‘z‘)fun1(); } return 0; }
标签：sim blank eof return har ros turn problem bsp
原文地址：http://www.cnblogs.com/liuqiyu/p/7290670.html
