搞ACM的前来看下，水题，但是WA

[vb2010]

题目在这里 杭电的 http://acm.hdu.edu.cn/showproblem.php?pid=2816
确实是水题 但是不知道为什么过不了 呢？

题目：
I Love You Too
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 516 Accepted Submission(s): 335



Problem Description
This is a true story. A man showed his love to a girl,but the girl didn't replied clearly ,just gave him a Morse Code:
****-/*----/----*/****-/****-/*----/---**/*----/****-/*----/-****/***--/****-/*----/----*/**---/-****/**---/**---/***--/--***/****-/ He was so anxious that he asked for help in the Internet and after one day a girl named "Pianyi angel" found the secret of this code. She translate this code as this five steps:
1.First translate the morse code to a number string:4194418141634192622374
2.Second she cut two number as one group 41 94 41 81 41 63 41 92 62 23 74,according to standard Mobile phone can get this alphabet:GZGTGOGXNCS

3.Third she change this alphabet according to the keyboard:QWERTYUIOPASDFGHJKLZXCVBNM = ABCDEFGHIJKLMNOPQRSTUVWXYZ
So ,we can get OTOEOIOUYVL
4.Fourth, divide this alphabet to two parts: OTOEOI and OUYVL, compose again.we will get OOTUOYEVOLI
5.Finally,reverse this alphabet the answer will appear : I LOVE YOU TOO

I guess you might worship Pianyi angel as me,so let's Orz her.
Now,the task is translate the number strings.


Input
A number string each line(length <= 1000). I ensure all input are legal.


Output
An upper alphabet string.


Sample Input
4194418141634192622374
41944181416341926223


Sample Output
ILOVEYOUTOO
VOYEUOOTIO


我的代码：

#include<iostream>
#include<string.h>
#define N 5
#define M 1010
using namespace std; 

int main()
{
	//定义变量
    char num[10][N];
    char c[M];
    char alp[M],a[M],b[M];
    int i,j,k,l,n,m,t;

	//数组初始化
    memset(c,0,sizeof(c));
    memset(alp,0,sizeof(alp));
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(num,0,sizeof(num));

	//打表(不知道是不是打表O(∩_∩)O~)
    num[2][1] = 'A', num[2][2] = 'B', num[2][3] = 'C';
    num[3][1] = 'D', num[3][2] = 'E', num[3][3] = 'F';
    num[4][1] = 'G', num[4][2] = 'H', num[4][3] = 'I';
    num[5][1] = 'J', num[5][2] = 'K', num[5][3] = 'L';
    num[6][1] = 'M', num[6][2] = 'N', num[6][3] = 'O';
    num[7][1] = 'P', num[7][2] = 'Q', num[7][3] = 'R', num[7][4] = 'S';
    num[8][1] = 'T', num[8][2] = 'U', num[8][3] = 'V';
    num[9][1] = 'W', num[9][2] = 'X', num[9][3] = 'Y', num[9][4] = 'Z';

	//输入数字字符串
    while(scanf("%s", &c) != EOF){
                      l = strlen(c);

					  //再次初始化
					  memset(alp,0,sizeof(alp));
					  memset(a,0,sizeof(a));
					  memset(b,0,sizeof(b));

					  //转换成标准手机键盘大写字母
                      for(i = 0; i < l+1 ; i+=2){
                            alp[i/2] = num[c[i] - '0'][c[i+1] - '0'];
                            }

					  //转换成对应的PC键盘字母
                      j = k = 0;
                      for(i = 0; i < l/2; i++){
                            switch(alp[i]){
                                           case 'Q' : alp[i] = 'A'; break;
                                           case 'W' : alp[i] = 'B'; break;
                                           case 'E' : alp[i] = 'C'; break;
                                           case 'R' : alp[i] = 'D'; break;
                                           case 'T' : alp[i] = 'E'; break;
                                           case 'Y' : alp[i] = 'F'; break;
                                           case 'U' : alp[i] = 'G'; break;
                                           case 'I' : alp[i] = 'H'; break;
                                           case 'O' : alp[i] = 'I'; break;
                                           case 'P' : alp[i] = 'J'; break;
                                           case 'A' : alp[i] = 'K'; break;
                                           case 'S' : alp[i] = 'L'; break;
                                           case 'D' : alp[i] = 'M'; break;
                                           case 'F' : alp[i] = 'N'; break;
                                           case 'G' : alp[i] = 'O'; break;
                                           case 'H' : alp[i] = 'P'; break;
                                           case 'J' : alp[i] = 'Q'; break;
                                           case 'K' : alp[i] = 'R'; break;
                                           case 'L' : alp[i] = 'S'; break;
                                           case 'Z' : alp[i] = 'T'; break;
                                           case 'X' : alp[i] = 'U'; break;
                                           case 'C' : alp[i] = 'V'; break;
                                           case 'V' : alp[i] = 'W'; break;
                                           case 'B' : alp[i] = 'X'; break;
                                           case 'N' : alp[i] = 'Y'; break;
                                           case 'M' : alp[i] = 'Z'; break;
                                           default : break; 
                                           }
							}

					  //将字符串截成前后两段
					  j = 0;
					  l = strlen(alp);
					  if(l % 2 != 0)
						  t = l / 2 + 1;
					  else
						  t = l / 2;
					  for(i = 0; i < t; i++){
							a[j] = alp[i];
							j++;
							}
					  k = 0;
					  for(i = t; i < l; i++){
							b[k] = alp[i];
							k++;
							}
					  j = k = 0;

					  //组合字符串
					  memset(alp,0,sizeof(alp));
					  for(i = 0; i < l; i+=2){
                                 alp[i] = a[j];
                                 alp[i + 1] = b[k];
								 j++;
								 k++;
                                 }
					  l = strlen(alp) - 1;

					  //逆序输出字符串
					  for(i = l; i >= 0; i--){
                            printf("%c",alp[i]);
                            }
                      printf("\n");
                      }
    
    //system("pause");
    return 0;
    } 

[我不是大明]

E文写的很乱,不要放链接，很多上CSDN的人打不开那些链接的。

[光宇广贞]

纯顶一个吧……

[beyond071]

mark 有空再来看

[LeonTown]

mark

[DarkChampion]

纯路过

[liao05050075]

while(scanf("%s", &c) != EOF)
读字符串的时候不用&,scanf("%s",c)就行了

[猫已经找不回了]

看看

[密斯刘]

[Quote=引用 1 楼 fz_zhou 的回复:]
E文写的很乱,不要放链接，很多上CSDN的人打不开那些链接的。
[/Quote]
打不开链接？？此话怎讲

[密斯刘]

[Quote=引用 2 楼 hikaliv 的回复:]
纯顶一个吧……
[/Quote]
thank you 我改用C++提交代码就OK了 原来用G++提交的代码

[密斯刘]

[Quote=引用 6 楼 liao05050075 的回复:]
while(scanf("%s", &c) != EOF)
读字符串的时候不用&,scanf("%s",c)就行了
[/Quote]

糊涂了好久了 谢谢提醒

[十八道胡同]

纯模拟就可以了

[dfkjsdhfks]

mark

[ToBeTough]

感觉可能用multimap<char,string>应该比较简单

[清海风缘]

路过

[密斯刘]

开始用G++提交的 出错
现在代码没改 用C++酒好了
不知道两者之间存在什么区别

[woaiweiyan]

为什么我这代码AC不了？
#include<stdio.h>
#include<string.h>
int main()
{
char c[1001],alp[1001],a[1001],b[1001];
char num[8][5]={"ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};
int i,j,k,l,n,m,t;
char s1[27]="QWERTYUIOPASDFGHJKLZXCVBNM",s2[27]="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
while(gets(c))
{
char *p,*q;
memset(alp,0,sizeof(alp));
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(i=0,p=c;*p;p++,i++)
alp[i]=num[*p-'0'-2][*(++p)-'0'-1];
for(i=0;alp[i];i++)
{
for(p=s1,q=s2;;p++,q++)
{
if(*p==alp[i])
{
alp[i]=*q;
break;
}
}
}
l=strlen(alp);
if(l%2!=0) t=l/2+1;
else t=l/2;
j=0;
for(i=0;i<t;i++)
a[j++]=alp[i];
j=0;
for(i=t;i<l;i++)
b[j++]=alp[i];
j=k=0;
memset(alp,0,sizeof(alp));
for(i=0;i<l;i+=2)
{
alp[i]=a[j++];
alp[i+1]=b[k++];
}
strrev(alp);
puts(alp);
}
return 0;
}