原题代号：HDU 1314
原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1314
Numerically Speaking
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 766 Accepted Submission(s):
190
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
...
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).
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.
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
（注意：数字是从第23列开始写，之前空格不能丢了）
解题思路：这个题类似二十六进制的互相转换，只不过a相当于1，z相当于26而已；只不过要考虑数字转换成字母时的特殊地方（26/26=1余0，但是26是z，所以如果余数为0则字母为z并且商-=1）
AC代码：
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL; //string比较函数:相等返回0，str1>str2返回1，str1<str2返回-1.
int Compare(string str1,string str2)
{
if(str1.length() > str2.length()) return ;
else if(str1.length() < str2.length()) return -;
else return str1.compare(str2);
} string Big_Plus(string str1,string str2)
{
string ans;
int len1=str1.length();
int len2=str2.length();
//将长度较小的前面补0，使两个string长度相同
if(len1<len2){
for(int i=;i<=len2-len1;i++){
str1=""+str1;
}
}else {
for(int i=;i<=len1-len2;i++){
str2=""+str2;
}
}
int len=max(len1,len2);
int carry=;
for(int i=len-;i>=;i--){
int tmp=str1[i]-''+str2[i]-''+carry;
carry=tmp/;
tmp%=;
ans=char(tmp+'')+ans;
}
if(carry) ans=char(carry+'')+ans;
return ans;
} //支持大数减小数
string Big_Sub(string str1,string str2)
{
string ans;
int carry=;
int difference=str1.length()-str2.length();//长度差
for(int i=str2.length()-;i>=;i--){
if(str1[difference+i]<str2[i]+carry){
ans=char(str1[difference+i]+-str2[i]-carry+'')+ans;
carry=;
}else {
ans=char(str1[difference+i]-str2[i]-carry+'')+ans;
carry=;
}
}
for(int i=difference-;i>=;i--){
if(str1[i]-carry>=''){
ans=char(str1[i]-carry)+ans;
carry=;
}else {
ans=char(str1[i]-carry+)+ans;
carry=;
}
}
//去除前导0
ans.erase(,ans.find_first_not_of(''));
if(ans.empty()) ans="";
return ans;
} string Big_Mul(string str1,string str2)
{
string ans;
int len1=str1.length();
int len2=str2.length();
for(int i=len2-;i>=;i--){
string tmpstr="";
int data=str2[i]-'';
int carry=;
if(data!=){
for(int j=;j<=len2--i;j++){
tmpstr+="";
}
for(int j=len1-;j>=;j--){
int t=data*(str1[j]-'')+carry;
carry=t/;
t%=;
tmpstr=char(t+'')+tmpstr;
}
if(carry!=) tmpstr=char(carry+'')+tmpstr;
}
ans=Big_Plus(ans,tmpstr);
}
ans.erase(,ans.find_first_not_of(''));
if(ans.empty()) ans="";
return ans;
} //正数相除，商为quotient,余数为residue void Big_Div(string str1,string str2,string& quotient,string& residue)
{
quotient=residue="";//商和余数清空
if(str2==""){//；判断除数是否为0
quotient=residue="ERROR";
return;
}
if(str1==""){//判断被除数是否为0
quotient=residue="";
return;
}
int res=Compare(str1,str2);
if(res<){//被除数小于除数
quotient="";
residue=str1;
return;
}else if(res==){
quotient="";
residue="";
return ;
}else {
int len1=str1.length();
int len2=str2.length();
string tmpstr;
tmpstr.append(str1,,len2-);//将str1的前len2位赋给tmpstr
for(int i=len2-;i<len1;i++){
tmpstr=tmpstr+str1[i];//被除数新补充一位
tmpstr.erase(,tmpstr.find_first_not_of(''));//去除前导0
if(tmpstr.empty()) tmpstr="";
for(char ch='';ch>='';ch--){//试商
string tmp,ans;
tmp=tmp+ch;
ans=Big_Mul(str2,tmp);//计算乘积
if(Compare(ans,tmpstr)<=){//试商成功
quotient=quotient+ch;
tmpstr=Big_Sub(tmpstr,ans);//减掉乘积
break;
}
}
}
residue=tmpstr;
}
quotient.erase(,quotient.find_first_not_of(''));
if(quotient.empty()) quotient="";
} string change(int num)
{
string n="";
stack<char>M;
while(num>)
{
M.push(num%+'');
num/=;
}
while(!M.empty())
{
n+=M.top();
M.pop();
}
return n;
} int change(string num)
{
int n=num[]-'';
for(int i=;i<num.size();i++)
n=n*+num[i]-'';
return n;
} int main()
{
//freopen("in.txt", "r", stdin);
string s;
while(cin>>s,s[]!='*')
{
if('a'<=s[]&&s[]<='z')
{
string num=change(s[]-'a'+);
for(int i=;i<s.size();i++)
{
num=Big_Plus(Big_Mul(num,change()),change(s[i]-'a'+));
}
cout<<s;
for(int i=s.size()+;i<;i++)cout<<' ';
for(int i=;i<num.size();i++)
{
if(i)
printf((num.size()-i)%==?",%c":"%c",num[i]);
else
printf("%c",num[i]);
}
cout<<endl;
}
else if(''<=s[]&&s[]<='')
{
stack<char>M;
string str1,str2;//商和余数
string str="",str3=s;
while(Compare(str3,"")>)//str1>0
{
Big_Div(str3,"",str1,str2);
str3=str1;
if(str2=="")
{
M.push('z');
str3=Big_Sub(str3,"");
}
else
{
M.push(change(str2)+'a'-);
}
}
while(!M.empty())
{
str+=M.top();
M.pop();
}
cout<<str;
for(int i=str.size()+;i<;i++)cout<<' ';
for(int i=;i<s.size();i++)
{
if(i)
printf((s.size()-i)%==?",%c":"%c",s[i]);
else
printf("%c",s[i]);
}
cout<<endl;
}
}
return ;
}