该题在ZOJ上的题号是1225。
题目描述如下：
Background
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, wor
本题不是难题，根据题意，只需把输入中的数字和字符串分开，然后分别按照相应的规则进行排序，并记录下第i个是数字或者是字符，最后按照记录情况一次输出相应的元素即可。因为需要对字符串数组进行排序，因此第一印象是使用C++的string和STL中的sort函数，但是结果却因为懒得写一个排序函数多写了很多代码。
具体代码如下：
#include < cstdio >
#include < cstdlib >
#include < string >
#include < cstring >
#include < algorithm >
#include < cctype >
using namespace std;
string s[ 80 ];
int a[ 80 ];
bool flag[ 80 ];
char buf[ 80 ];
bool cmp( string a, string b)
{
string tmpa = a;
string tmpb = b;
transform(a.begin(),a.end(), tmpa.begin(), ::tolower);
transform(b.begin(),b.end(), tmpb.begin(), ::tolower);
return tmpa < tmpb;
}
int main( void )
{
int alpha, num;
string t;
char * p;
int i, j;
int sign;
int tmp;
int k;
int m, n;
freopen( " in.txt " , " r " , stdin);
while (fgets(buf, 80 , stdin) != NULL && buf[ 0 ] != ' . ' )
{
p = buf;
i = j = 0 ;
alpha = num = 0 ;
k = 0 ;
while ( * p != ' \n ' && * p != ' . ' )
{
while ( * p == ' ' )
++ p;
sign = 1 ;
if ( * p == ' - ' || isdigit( * p) || * p == ' + ' )
{
if ( * p == ' - ' )
{
sign = - 1 ;
++ p;
}
else if ( * p == ' + ' )
++ p;
tmp = 0 ;
while ( * p != ' , ' && * p != ' . ' )
{
tmp = tmp * 10 + ( * p - ' 0 ' );
++ p;
}
a[num ++ ] = tmp * sign;
flag[k ++ ] = false ;
}
else
{
i = 0 ;
t = "" ;
while ( * p != ' , ' && * p != ' . ' )
{
t += * p;
++ p;
++ i;
}
s[alpha ++ ] = t;
flag[k ++ ] = true ;
}
++ p;
}
sort(a, a + num);
sort(s, s + alpha, cmp);
m = n = 0 ;
if ( ! flag[ 0 ])
{
printf( " %d " , a[ 0 ]);
++ m;
}
else
{
printf( " %s " , s[ 0 ].c_str());
++ n;
}
for (i = 1 ; i < k; ++ i)
{
if ( ! flag[i])
{
printf( " , %d " , a[m]);
++ m;
}
else
{
printf( " , %s " , s[n].c_str());
++ n;
}
}
printf( " .\n " );
}
return 0 ;
}