sgu 460. Plural Form of Nouns

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最新推荐文章于 2020-11-05 00:18:52 发布

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本文介绍了一个简单的程序，用于将英语名词从单数形式转换为复数形式。该程序遵循特定的转换规则，例如以 ch、x、s 或 o 结尾的单词加 es，以 f 或 fe 结尾的单词变为 ves，以 y 结尾的单词变为 ies，其余情况直接加 s。文章提供了完整的 C++ 实现代码。

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460. Plural Form of Nouns

Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

In the English language, nouns are inflected by grammatical number — that is singular or plural. In this problem we use a simple model of constructing plural from a singular form. This model doesn't always make English plural forms correctly, but it works in most cases. Forget about the real rules you know while solving the problem and use the statement as a formal document.

You are given several nouns in a singular form and your program should translate them into plural form using the following rules:

If a singular noun ends with
```
ch
```
,
```
x
```
,
```
s
```
,
```
o
```
the plural is formed by adding
```
es
```
. For example,
```
witch
```
->
```
witches
```
,
```
tomato
```
->
```
tomatoes
```
.
If a singular noun ends with
```
f
```
or
```
fe
```
, the plural form ends with
```
ves
```
. For example,
```
leaf
```
->
```
leaves
```
,
```
knife
```
->
```
knives
```
. Pay attention to the letter
```
f
```
becoming
```
v
```
.
Nouns ending with
```
y
```
change the ending to
```
ies
```
in plural. For example,
```
family
```
->
```
families
```
.
In all other cases plural is formed by adding
```
s
```
. For example,
```
book
```
->
```
books
```
.

Input

The first line of input contains a single positive integer n (1 ≤ n ≤ 10) — the number of words to be processed. The following n lines contain one word each. A word consists from 2 to 25 lowercase Latin letters. It is not guaranteed that the given words are real English words from vocabulary.

Output

Print n given words in their plural forms on separate lines. Keep the words in the same order as they are given in the input.

Example(s)

sample input	sample output
3 contest hero lady	contests heroes ladies

变复数形式

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

char a[30];

void solve(){
    int n;
    scanf("%d",&n);
    while (n--){
        scanf("%s",a);
        int len=strlen(a);
        if(a[len-1]=='o'||a[len-1]=='s'||a[len-1]=='x'||(a[len-2]=='c'&&a[len-1]=='h')){
            a[len]='e';
            a[len+1]='s';
            a[len+2]='\0';
        }
        else if(a[len-1]=='f'){
            a[len-1]='v';
            a[len]='e';
            a[len+1]='s';
            a[len+2]='\0';
        }
        else if(a[len-2]=='f'&&a[len-1]=='e'){
            a[len-2]='v';
            a[len-1]='e';
            a[len]='s';
            a[len+1]='\0';
        }
        else if(a[len-1]=='y'){
            a[len-1]='i';
            a[len]='e';
            a[len+1]='s';
            a[len+2]='\0';
        }
        else{
            a[len]='s';
            a[len+1]='\0';
        }

        printf("%s\n",a);
    }
}




int main() {


#ifdef LOCAL_DEFINE
    freopen("input.txt", "rt", stdin);
#endif

    solve();


    return 0;
}