//MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)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(jié)的一個(gè)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(kāi)始是用的庫(kù)函數(shù) strtok 對(duì)字符串進(jìn)行處理, 直接敲完,沒(méi)有出現(xiàn)錯(cuò)誤, 提交,悲劇開(kāi)始了
下面的是
PE 的代碼 :
//MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)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?yàn)閟trtok是忽視被截字符串的個(gè)數(shù)的 "-----######---##-#-#"
這組數(shù)據(jù)應(yīng)該輸出10個(gè)空格, 而我的代碼值能輸出3個(gè).
下面的是
AC代碼 :
//MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)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(gè)下午加一個(gè)晚上才 AC , 是自己把簡(jiǎn)單問(wèn)題想太復(fù)雜了............
Roowe 神牛代碼 :
//MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)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;
}