发短信
时间限制:1000 ms | 内存限制:65535 KB
难度:3
- 描述
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下图是手机常用的九键英文输入法界面,如果要输入字母'A',我们只需要按一次数字键2,按键顺序记为2;如果要输入字母'B'的话,我们需要连续按两次数字键2,按键顺序记为22;同理:字母'C’需要连续按3次数字键2,按键顺序记为222。通过这种方法,我们用手机10多个键就能输入26个英文字母。现在你的任务是统计一段英文用手机输入的按键顺序,同样,你也要能把按键顺序翻译成相应的英文内容。为了使问题简化,我们假设内容只有大写英文字母和空格。输入 有多组测试数据
每组测试数据占一行,有两种情况:
(1)短信内容(只含有若干个空格和大写字母,不超过1000个字符)
(2)短信按键顺序(只含有若干空格和数字,其中第一个肯定是数字,不超过1000个字符) 输出 对于每组测试数据:
如果是短信内容,输出每个字母的按键顺序,每个字母的按键顺序用空格隔开
如果是按键顺序,输出它代表的内容样例输入I LOVE YOU HELLO WORLD 444 0 555 666 888 33 0 999 666 88 44 33 555 555 666 0 9 666 777 555 3
样例输出444 0 555 666 888 33 0 999 666 88 44 33 555 555 666 0 9 666 777 555 3 I LOVE YOU HELLO WORLD
编译代码:刚学c++,借用的库:#pragma warning(disable:4786) #include<iostream> #include<stdio.h> #include<string> #include<cstring> #include<map> using namespace std; map<string,string>m; map< string , string >::iterator it ; void init() { m["0"] = " " ; m["2"] = "A" ; m["22"] = "B" ; m["222"] = "C" ; m["3"] = "D" ; m["33"] = "E" ; m["333"] = "F" ; m["4"] = "G" ; m["44"] = "H" ; m["444"] = "I" ; m["5"] = "J" ; m["55"] = "K" ; m["555"] = "L" ; m["6"] = "M" ; m["66"] = "N" ; m["666"] = "O" ; m["7"] = "P" ; m["77"] = "Q" ; m["777"] = "R" ; m["7777"] = "S" ; m["8"] = "T" ; m["88"] = "U" ; m["888"] = "V" ; m["9"] = "W" ; m["99"] = "X" ; m["999"] = "Y" ; m["9999"] = "Z" ; m[" "] = "0" ; m["A"] = "2" ; m["B"] = "22" ; m["C"] = "222" ; m["D"] = "3" ; m["E"] = "33" ; m["F"] = "333" ; m["G"] = "4" ; m["H"] = "44" ; m["I"] = "444" ; m["J"] = "5" ; m["K"] = "55" ; m["L"] = "555" ; m["M"] = "6" ; m["N"] = "66" ; m["O"] = "666" ; m["P"] = "7" ; m["Q"] = "77" ; m["R"] = "777" ; m["S"] = "7777" ; m["T"] = "8" ; m["U"] = "88" ; m["V"] = "888" ; m["W"] = "9" ; m["X"] = "99" ; m["Y"] = "999" ; m["Z"] = "9999" ; } void fun(char a[]) { it = m.find(a) ; if( it != m.end() ) cout<<it->second<<" "; } void fuu(char a[]) { it = m.find(a) ; if( it != m.end() ) cout<<it->second; } int main() { char str[1000],s[10]; init(); int i=1,j,k,n; while( gets(str)!=NULL) {k=0; if(str[0]>='0' && str[0]<='9')//判断输入的是数字还是字符 {n=0; int x=strlen(str);//把接收到的字符串都转化成单个的短字符串 for(i=0; i<x; i++) { if(str[i]!=' ') { s[n++]=str[i]; if(i==x-1) {s[n]='