poj1520scramble_sort

Scramble Sort

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 1896 Accepted: 972

Description

In this problem you will be given a series of lists containing both words and numbers. The goal is to sort these lists in such a way that all words are in alphabetical order and all numbers are in numerical order. Furthermore, if the nth element in the list is a number it must remain a number, and if it is a word it must remain a word.

Input

The input will contain multiple lists, one per line. Each element of the list will be separated by a comma followed a space, and the list will be terminated by a period. The input will be terminated by a line containing only a single period.

Output

For each list in the input, output the scramble sorted list, separating each element of the list with a comma followed by a space, and ending the list with a period.

Sample Input

0.banana, strawberry, OrAnGe.Banana, StRaWbErRy, orange.10, 8, 6, 4, 2, 0.x, 30, -20, z, 1000, 1, Y.50, 7, kitten, puppy, 2, orangutan, 52, -100, bird, worm, 7, beetle..

Sample Output

0.banana, OrAnGe, strawberry.Banana, orange, StRaWbErRy.0, 2, 4, 6, 8, 10.x, -20, 1, Y, 30, 1000, z.-100, 2, beetle, bird, 7, kitten, 7, 50, orangutan, puppy, 52, worm.

// poj1028.cpp : 定义控制台应用程序的入口点。//#include <stdio.h>#include <iostream>#include <string>using namespace std;int flag[100];//标志原串中是数字还是单词string word[30];//输入存储数组int dig[30];//保存数字string cha[30];//保存单词char str[100];//输入存储数组int i,j,k,len,clen,dlen;//int input(){i=0;cin>>str;//输入字符串if(str[0]=='.')//若是一个“.”，结束return 1;while(str[strlen(str)-1]!='.')//考虑进行改进{str[strlen(str)-1]= '/0';word[i]=str;if(isalpha(str[0]))//判断是否是字母flag[i]=1;elseflag[i]=0;i++;cin>>str;}str[strlen(str)-1]='/0';word[i]=str;if(isalpha(str[0]))flag[i]=1;elseflag[i]=0;len=i+1;//记录输入的数字和单词的总个数return 0;}int atoi(string s,int i)//将ASCII码数字转换成int型{int n=0;for(;i<s.length()&&s[i]>='0'&&s[i]<='9';++i)n=10*n+(s[i]-'0');return n;}void partition()//将数字和单词分开保存的函数{j=0,k=0,clen=0,dlen=0;for(i=0;i<len;i++){if(flag[i]==1)//字母{cha[j]=word[i];j++;clen++;//记录有多少个单词}else{if(word[i][0]!='-')dig[k]=atoi(word[i],0);else{dig[k]=(-1)*atoi(word[i],1);}k++;dlen++;//记录有多少个数字}}}int cmp2(string a,string b)//用于单词的比较{char c[30];//用于临时保存被小写化得单词(题目中要求不区分大小写)char d[30];for(int i=0;i<a.length();i++){c[i]=isupper(a[i]) ? a[i]-'A' +'a':a[i];}for(int i=0;i<b.length();i++){d[i]=isupper(b[i]) ? b[i]-'A' +'a':b[i];}if(strcmp(c,d)<=0)return 1;elsereturn 0;}int part1(string a[],int p,int r)//单词数组的quick_sort中的partition()函数{string x=a[r];int i=p-1;string temp;for(int j=p;j<=r-1;j++){if(cmp2(a[j],x)){i++;temp=a[j];a[j]=a[i];a[i]=temp;}}temp=a[r];a[r]=a[i+1];a[i+1]=temp;return i+1;}int part2(int a[],int p,int r)//数字数组的quick_sort中的partition()函数{int x=a[r];int i=p-1;int temp;for(int j=p;j<=r-1;j++){if(a[j]<=x){i++;temp=a[j];a[j]=a[i];a[i]=temp;}}temp=a[r];a[r]=a[i+1];a[i+1]=temp;return i+1;}void quicksort2(int a[],int p,int r){if(p<r){int q=part2(a,p,r);quicksort2(a,p,q-1);quicksort2(a,q+1,r);}}void quicksort1(string a[],int p,int r){if(p<r){int q=part1(a,p,r);quicksort1(a,p,q-1);quicksort1(a,q+1,r);}}void output(){for(int i=0,j=0,k=0;i<len;i++){if(i!=len-1){if(flag[i]==1)//字符串{cout<<cha[j]<<", ";j++;}else{cout<<dig[k]<<", ";k++;}}else{if(flag[i]==1)//字符串{cout<<cha[j]<<'.';j++;}else{cout<<dig[k]<<'.';k++;}}}cout<<'/n';}int main(void){while(1){if(input())break;partition();//将单词和数字分开quicksort1(cha,0,clen-1);quicksort2(dig,0,dlen-1);output();}return 0;}