Posted on 2010-08-04 20:04
MiYu 閱讀(521)
評論(2) 編輯 收藏 引用 所屬分類:
C/C++ 、
ACM ( 模擬 ) 、
ACM ( 串 ) 、
ACM ( 水題 )
//MiYu原創, 轉帖請注明 : 轉載自 ______________白白の屋
題目地址 :
http://acm.hdu.edu.cn/showproblem.php?pid=1982
PE了N次, 很糾結的一個題........ 題目如下 :
Problem Description
Do you know Kaitou Kid? In the legend, Kaitou Kid is a master of disguise, and can take on the voice and form of anyone. He is not an evil person, but he is on the wrong side of the law. He's the very elusive phantom thief who never miss his prey although he always uses word puzzles to announce his targets before action.
You are the leader of a museum. Recently, you get several priceless jewels and plan to hold an exhibition. But at the moment, you receive Kid's word puzzle... Fortunately, It seems Kid doesn’t want to trouble you, and his puzzle is very easy. Just a few minutes, You have found the way to solve the puzzle:
(1) change 1 to 'A', 2 TO 'B',..,26 TO 'Z'
(2) change '#' to a blank
(3) ignore the '-' symbol, it just used to separate the numbers in the puzzle
Input
The first line of the input contains an integer C which means the number of test cases. Then C lines follow. Each line is a sentence of Kid’s word puzzle which is consisted of '0' ~ '9' , '-' and '#'. The length of each sentence is no longer than 10000.
Output
For each case, output the translated text.
Sample Input
4
9#23-9-12-12#19-20-5-1-12#1-20#12-5-1-19-20#15-14-5#10-5-23-5-12
1-14-4#12-5-1-22-5#20-8-5#13-21-19-5-21-13#9-14#20#13-9-14-21-20-5-19
1-6-20-5-18#20-8-5#15-16-5-14-9-14-7#15-6#20-8-5#5-24-8-9-2-9-20-9-15-14
7-15-15-4#12-21-3-11
Sample Output
I WILL STEAL AT LEAST ONE JEWEL
AND LEAVE THE MUSEUM IN T MINUTES
AFTER THE OPENING OF THE EXHIBITION
GOOD LUCK
剛開始是用的庫函數 strtok 對字符串進行處理, 直接敲完,沒有出現錯誤, 提交,悲劇開始了
下面的是
PE 的代碼 :
//MiYu原創, 轉帖請注明 : 轉載自 ______________白白の屋
#include <iostream>
#include <cstdlib>
#include <cstring>
using namespace std;
char psw[10005];
char sym[133];
void setSym ( )
{
int i;
char ch;
for ( ch = 'A', i = 1; i <= 26; ++ i , ++ ch )
{
sym[i] = ch ;
}
}
string prs ( char *psw )
{
string str;
int n = strlen ( psw );
int num = 0;
for ( int i = 0; i != n; ++ i )
{
if ( psw[i] != '-' )
{
num = num * 10 + psw[i] - '0';
}
else
{
if ( num != 0 )
{
str += sym[num];
}
num = 0;
}
}
if ( num != 0 )
{
str += sym[num];
}
return str;
}
int main ()
{
setSym ();
int T;
while ( scanf ( "%d\n",&T ) != EOF )
{
while ( T -- )
{
gets ( psw );
string str;
char *ptr = strtok ( psw, "#" );
if ( strlen ( ptr ) != 0 )
str = prs ( ptr );
while ( ptr = strtok ( NULL, "#" ) )
{
if ( strcmp ( ptr, "" ) != 0 )
{
str += " ";
str += prs ( ptr );
}
}
cout << str << endl;;
}
}
return 0;
}
最后在Ambition 大牛的提示下,成功AC, 因為strtok是忽視被截字符串的個數的 "-----######---##-#-#"
這組數據應該輸出10個空格, 而我的代碼值能輸出3個.
下面的是
AC代碼 :
//MiYu原創, 轉帖請注明 : 轉載自 ______________白白の屋
#include <iostream>
#include <cstdlib>
#include <cstring>
using namespace std;
char psw[20005];
char temp[20005];
char sym[133];
void setSym ( )
{
int i;
char ch;
for ( ch = 'A', i = 1; i <= 26; ++ i , ++ ch )
{
sym[i] = ch ;
}
}
string prs ( char *psw )
{
string str;
int n = strlen ( psw );
int num = 0;
for ( int i = 0; i != n; ++ i )
{
if ( psw[i] != '-' )
{
num = num * 10 + psw[i] - '0';
}
else
{
if ( num != 0 )
{
str += sym[num];
}
num = 0;
}
}
if ( num != 0 )
{
str += sym[num];
}
return str;
}
int main ()
{
setSym ();
int T;
while ( scanf ( "%d",&T ) != EOF )
{
getchar ();
while ( T -- )
{
gets ( psw );
int len = strlen ( psw );
int beg = 0;
memset ( temp, '\0', sizeof ( temp ) );
string str;
while ( psw[beg] != '\0' )
{
int i = 0;
while ( psw[beg] != '#' && psw[beg] != '\0' )
{
temp[i++] = psw[beg++];
}
temp[i] = '\0';
string t = prs ( temp );
if ( t.size() != 0 )
{
str += t;
if ( psw[beg] == '#' )
{
str += " ";
}
}
else if ( psw[beg] == '#' )
{
str += " ";
}
beg ++;
}
cout << str << endl;
memset ( psw, '\0', sizeof ( psw ) );
}
}
return 0;
}
弄了一個下午加一個晚上才 AC , 是自己把簡單問題想太復雜了............
Roowe 神牛代碼 :
//MiYu原創, 轉帖請注明 : 轉載自 ______________白白の屋
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <ctype.h>
using namespace std;
char str[27]={"ABCDEFGHIJKLMNOPQRSTUVWXYZ"};
char s[10001];
int main()
{
int T,len,i,num;
scanf("%d",&T);
getchar();
while(T--)
{
gets(s);
len=strlen(s);
for(i=0;i<len;i++)
{
if(isdigit(s[i]) && isdigit(s[i+1]))
{
num=10*(s[i]-'0')+s[i+1]-'0';
printf("%c",str[num-1]);
i++;
continue;
}
if(isdigit(s[i]) && !isdigit(s[i+1]))
{
num=s[i]-'0';
printf("%c",str[num-1]);
continue;
}
if(s[i]=='#') printf(" ");
}
printf("\n");
}
return 0;
}