描述
如果你计划读一本英语小说,可能你需要做一些提前准备,例如列出书中“常见的生词”。也许在以前,这是一件不可能的事。因为你既然没读过这本书,又怎么知道哪些是常见的词呢。但是随着计算机和电子书的普及,这边为了可能,在用计算机统计完一本书后,你可能发现这样的结果:排在前几位的都是一些非常简单的词语——the(3626次)、and(1922次)、to(1860次)、he(1748次)、a(1691次)。而在你翻过一两百个单词后就会开始出现一些你想要的东西。
现在从文章中提取单词的工作已经做好了,你需要做的就是统计一下排名。每个单词的排名=出现次数更高的单词个数+1,所有单词全部转换为小写形式进行统计和输出,所有出现次数一样的单词排名一样,并且可能输出不止n个单词。排列上排名相同的单词以字典序列出。
输入
第1行:一个整数n(1<n<=10000),代表输出的最大排名
第2行:N个单词,每个单词后跟一个空格,单词仅由字母组成,最大长度50,最大不同单词个数10000
本题不显示给出N,以EOF判断输入结束
输出
第M行:Rank r: word (count),r为单词排名,如果为个位数,左边多空一格,排名后接一个冒号一个空格,word为单词,单词后再加一个空格,count为出现次数,用小括号括起来。
样例输入
15
Albus Rose Hugo and Lily laughed The train began to move and Harry walked alongside it watching his son s thin face already ablaze with excitement Harry kept smiling and waving even though it was like a little bereavement watching his son glide away from him The last trace of steam evaporated in the autumn air The train rounded a corner Harry s hand was still raised in farewell He ll be alright murmured Ginny As Harry looked at her he lowered his hand absentmindedly and touched the lightning scar on his forehead I know he will The scar had not pained Harry for nineteen years All was well
样例输出
Rank 1: the (6)
Rank 2: harry (5)
Rank 3: and (4)
Rank 3: his (4)
Rank 5: he (3)
Rank 5: was (3)
Rank 7: a (2)
Rank 7: hand (2)
Rank 7: in (2)
Rank 7: it (2)
Rank 7: s (2)
Rank 7: scar (2)
Rank 7: son (2)
Rank 7: train (2)
Rank 7: watching (2)
题目来源
这题输入和输出有点困惑,之后看了学长的代码才懂得,花了我一个半小时才AC掉,值得我收获的是对排序函数(sort)又有了新的认识及用法
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct stu
{
char ch[50];
int sum;
}a[10000];
int n;
int cmp(stu a,stu b)
{
if(a.sum>b.sum)
return 1;
else if(a.sum==b.sum)
{
if(strcmp(a.ch,b.ch)<0)
return 1;
else
return 0;
}
else
return 0;
}
int main()
{
char c,ch[10000][50],b[50]="++";
while(cin>>n)
{
int i=0,k=0,j=0,s=0;
getchar();
while((c=getchar())!='\n')
{
if(c>='A' && c<='Z')
c+=32;
if(c>='a' && c<='z')
{
ch[k][j]=c;
j++;
}
if(c==' ')
{
k++;
j=0;
}
}
while(i<=k)
{
if(strcmp(ch[s],b)!=0)
{
a[i].sum=1;
strcpy(a[i].ch,ch[s]);
for(j=s+1;j<=k;j++)
{
if(strcmp(ch[j],a[i].ch)==0)
{
strcpy(ch[j],b);
a[i].sum++;
}
}
i++;
}
s++;
}
sort(a+0,a+k+1,cmp);j=1;
printf("Rank %d: %s (%d)\n",j,a[0].ch,a[0].sum);
for(i=1;i<n;i++)
{
if(a[i].sum!=a[i-1].sum)
j=i+1;
printf("Rank %d: %s (%d)\n",j,a[i].ch,a[i].sum);
}
}
return 0;
}
标签:ch,++,sum,Rank,单词,his,简单 From: https://blog.51cto.com/u_15896805/5897489