1.     项目需求：1、分析文本文件的每个词语出现的频率；

2.                      2、将最高频率的10个词打印出来。

3.     规格说明：要求文本文件在30k~300K之间。

4.     概要设计：将文本文件读取出来-〉读取过程中将每个词记录下来，并计数-〉找出计数最多的10个词输出

5.     实现步骤：

     #include "stdio.h"

#include "stdlib.h"

#include "ctype.h"

#include "string.h"

struct L{

    char p[30];

    int n;

};

int sum=0;

void read(struct L word[]);

void sort(struct L word[]);

void main(){

    struct L word;

    char *a[50]={"i","the","and","a","to","of","he","you","it",

    "we","us","after","there","are","is","from","me","them","about",

    "was","were","on","in","your","that","all","him","his","at","up",

    "out","but","for","this","didn't","very","or","with","over","around"

    ,"had","just","","have","too","always"};

    read(word);

    sort(word);

    while(!strcmp(word.p,a)){

        cout<<word.p<<endl);

    }

    cout<<"总运行时间:"<<(finish-start)/1000<<"s"<<endl;

}

}

void read(struct L word[])

{

    ifstream in("d:/try.txt");

    in>>noskipws;

    if(!in) {cout<<"cannot open!"<<endl;return;}

    char ch,temp[30];

    while(in)

    {

        int i=0;

        in>>ch;

        temp[0]='\0';

        while((ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z')||temp[0]=='\0')

        {

            if(ch>='a'&&ch<='z'||ch>='A'&&ch<='Z')

            {

                temp[i]=ch;

                i++;

            }

            in>>ch;

            if(in.eof())break;

           

        }

        temp[i]='\0';

        for(i=0;i

        {

            if(!_stricmp(temp,word[i].p))

            { word[i].n++;break;}

        }

        if(i==sum)

        {

                strcpy(word[sum].p,temp);

                word[sum].n=1;

                    sum++;

 

 

        }

    }

    cout<<"读文件,分出单词并统计的时间:"<<(f-s)/1000<<"s"<<endl;

    in.close();

}

void sort(struct L word[])

{

    struct L temp;

    for(int i=0;i

        for(int j=0;j

            if(word[j].n

            {

                strcpy(temp.p,word[j].p);

                temp.n=word[j].n;

                strcpy(word[j].p,word[j+1].p);

                word[j].n=word[j+1].n;

                strcpy(word[j+1].p,temp.p);

                word[j+1].n=temp.n;

            }

}

http://s5/mw690/003i71TNgy6NBKIeHWY44&690

性能分析：http://s16/mw690/001YxvFkgy6HsPAbZHp7f&690

http://s9/mw690/001YxvFkgy6HsPAgNxCc8&690

http://s8/mw690/004awxmYgy6MZ7sZMeX97&690
http://s3/mw690/004awxmYgy6MZ7xgDJg82&690

http://s5/mw690/004awxmYgy6MZ7xk8G824&690

从图中可以看出，main函数的已用非独占时间百分比几乎占用程序运行的整个事件，而其本身的已用独占时间百分比只占15%，执行main函数本身并不占用太多时间。read函数的已用非独占时间百分比为85%，是程序主要实现功能的部分，但其独占时间百分比仅为4.6%，而strcmp函数是read函数中实现本程序核心功能的函数，每读取一个单词将与前面已存储的单词依次进行比较，已用独占时间百分比为60%。