以上图片来自新浪微博。
本题要求你实现一个稍微更值钱一点的 AI 英文问答程序,规则是:
I
;can you
、could you
对应地换成 I can
、I could
—— 这里“独立”是指被空格或标点符号分隔开的单词;I
和 me
换成 you
;?
换成惊叹号 !
;输入首先在第一行给出不超过 10 的正整数 N,随后 N 行,每行给出一句不超过 1000 个字符的、以回车结尾的用户的对话,对话为非空字符串,仅包括字母、数字、空格、可见的半角标点符号。
按题面要求输出,每个 AI 的回答前要加上 AI:
和一个空格。
6
Hello ?
Good to chat with you
can you speak Chinese?
Really?
Could you show me 5
What Is this prime? I,don 't know
Hello ?
AI: hello!
Good to chat with you
AI: good to chat with you
can you speak Chinese?
AI: I can speak chinese!
Really?
AI: really!
Could you show me 5
AI: I could show you 5
What Is this prime? I,don 't know
AI: what Is this prime! you,don't know
有两次 if(s[len-1]==' ')s.replace(len-1,1,""); 的原因是使用replace函数会产生莫名的字符串内容的错误 如图
第一次去除前后总共两个空格后末尾依旧会有一个空格,所以再次去除末尾空格!
此代码在编译器运行会报警告 warning: comparison between signed and unsigned integer expressions [-Wsign-compare]| ,不需要将该行的int改为unsigned int ,否则pta平台会报段错误
代码如下:
#include
using namespace std;
string s;
string s1[]={"can you","could you","I","me"};
string s2[]={"@","#","you","you"}; /*使用符号作为中间替代,避免将"can you"替换为"I can"之后"I"被替换为"you"*/
string s3[]={"@","#","?"};
string s4[]={"I can","I could","!"};
void trim(){
int pos=s.find(" ");
while(pos!=-1){ //将字符串所有的大于等于两个空格的替换为一个空格
s.replace(pos,2," ");
pos=s.find(" ");
}
//find()函数在字符串中寻找到括号中子字符串的下标并返回,失败返回-1
int len=s.length();
if(s[0]==' ') //去除字符串首部空格
s.replace(0,1,"");
if(s[len-1]==' ') //去除字符串末尾空格
s.replace(len-1,1,"");
len=s.length();
for(int i=len-1;i>=0;i--){ //将标点符号前面的空格去除
if(s[i-1]==' ' && ispunct(s[i]))
s.replace(i-1,1,"");
}
//ispunct()函数判断括号内字符是否为标点符号,标点符号返回true,否则返回false
len=s.length();
if(s[len-1]==' ') //去除字符串末尾空格
s.replace(len-1,1,"");
}
void change(){
for(int i=0;i<4;i++) //将"can you","could you","I","me"替换为"@","#","you","you"
{
int leni=s1[i].length();
for(int j=0; j>n;
getchar();
while(n--){
getline(cin,s);
cout<