Question

c语言编程，请高手帮忙！有最高加分！！！200

用C语言编写软件完成以下任务：输入一行字符存入文件myfile.txt中，读取文件数据后统计字符的个数，统计喊有某一字串的个数。例如：在abxyefghabcd中有ab子串2个。

Answer 1

我做到了，希望楼主也加到满分哟，呵呵。

程序在找子串时，如在abcccc中找子串cc时结果是3个而不是2个，当然楼主如果要求是2个稍改下就成。

输入字符串时允许空格。

用法：

1 输入一串字符，如123 abvc uuuu(回车）

2 程序显示统计有多少字符的结果，并要求输入想查的子串此时输入想查的子串，如 uu（回车）

3 程序显示有3个uu子串在里边。

注：我规定的是只有字母才算字符。
myfile.txt生成在D盘上。
编译环境TC2.0

#include"stdio.h"
#include"stdlib.h"
void main()
{

FILE *fp;

char ch,*p,*pcopy,*pp,a[1000],*tiaojian;

int i=0;

if((fp=fopen("d:\\myfile.txt","w+"))==NULL)
{
printf("Cannot open file strike any key exit!");
getch();
exit(1);
}

printf("please input the sentence here : \n");

gets(p);

while(*p!='\0') {

ch=*p++;

fputc(ch,fp);} /*写到myfile.txt*/

fclose(fp);

if((fp=fopen("d:\\myfile.txt","r+"))==NULL) /*从myfile.txt中读取文件*/
{
printf("Cannot open file strike any key exit!");
getch();
exit(1);
}

p=a;

while ((*p++=fgetc(fp))!=EOF) {

if ((*(p-1)<='z' && *(p-1)>='a') || (*(p-1)<='Z' && *(p-1)>='A')) i++;}

*p='\0';

printf("\n\nCOUNT:there are %d words in your sentence!",i);

p=a;

printf("\n\ninput the words that you want to look for: \n");

fflush(stdin);

gets(tiaojian);

pp=tiaojian;

i=0;

while (*p!='\0') {

pcopy=p++;

while (*pcopy++==*tiaojian++) {if (*tiaojian=='\0') {i++;break;}}

tiaojian=pp; }

printf("\n\n%d ge zi chuan!",i);

fclose(fp);

getch();
}

Answer 2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int GetSubstrNum(char* S, char* T);

int main(int argc, char* argv[])
{
char T[81];
FILE *infp, *outfp;
if(!(infp = fopen("myfile.txt", "rt"))){
printf("error on open myfile.txt file!\n");
getchar();
exit(1);
}
if(!(outfp = fopen("Statistic.txt","wt"))){
printf("error on open Statistic.txt file!\n");
getchar();
exit(1);
}
printf("请输入要寻找的子串：");
scanf("%s",T);
int count = 0;
while(!feof(infp)){
char str[81];
if(fscanf(infp, "%s", str) == EOF)
break;
for(int index = 0; str[index] >='a' && str[index] <= 'z'; index++){
count++;
}
fprintf(outfp,"%s", str);
fprintf(outfp, " 字符串个数 %d",count);
fprintf(outfp, " 相同子串%s数 %d\n", T, GetSubstrNum(str, T));

}

fclose(outfp);
fclose(infp);
return 0;
}

int GetSubstrNum(char* S, char* T)
{
if(strlen(S)== 0 || strlen(T) == 0)
return 0;
int i = 0, j = 0;
int count = 0;
while(i < strlen(S)){
if(S[i] == T[j]){
++i;
++j;
}
else{
i = i-j+1;
j = 0;
}
if(j >= strlen(T)) {
count++;
j = 0;
}
}
return count;
}
示例：
asdafasldfasldkfjalskdjflasdjflksdfjlasdjflas 字符串个数 45 相同子串sdf数 1

Answer 3

#include<stdio.h>
#include<stdlib.h>
void main()
{
char one_line[200];
char str[80],sub_str[10];
int i,j,gb,len,l,n;
int total_byte;
FILE *fout,*fin;

// write 3 line letters
fout = fopen("myfile.txt","w");
for (i=0;i<3;i++){
printf("Enter one line letters\n");
gets(&one_line[0]);
fprintf(fout,"%s\n",one_line);
};
fclose(fout);

// open file
fin= fopen("myfile.txt","r");
// how many chars
while ( gb = fgetc(fin) !=EOF ) { total_byte = total_byte + 1;};
rewind(fin);
printf("total_byte: %d\n",total_byte);

// how many sub_str

printf("Enter sub string ab\n");
scanf("%s",&sub_str[0]);
l=strlen(sub_str);
n=0;

while (fscanf(fin,"%s",&str[0]) !=EOF){
len = strlen(str);
if (l <= len){
for (i=0;i<=len-1;i++){
if ( strncmp(&str[0]+i,sub_str,l) == 0) n = n + 1;
}
}
}
fclose(fin);
printf("n=%d\n",n);
getch();
}

Answer 4

#include<stdio.h>
#include<string.h>
void main()
{
FILE *fp;
char ch1[100],ch2[100];
int i,j,k,a,b,count=0;
printf("input the string:");
scanf("%s",ch1);
if((fp=fopen("F:\\myfile.txt","w"))==NULL)
{
printf("Can not open the file!\n");
exit(0);
}
j=strlen(ch1);
fwrite(ch1,j,1,fp);
fclose(fp);
printf("input the baby string:");
scanf("%s",ch2);
k=strlen(ch2);
if(k>j)
{
printf("input again:");
scanf("%s",ch2);
}
for(i=0;i<j-k+1;i++)
{
a=i;
b=0;
while((ch1[a]==ch2[b]) && a<=k+i-1)
{
a++;
b++;
}
if(a-1==k+i-1)
count++;
}

printf("%d\n",count);
}

Answer 5

用这个看嘛
#include<stdio.h>
#include<string.h>
void main()
{
FILE *fp;
char ch1[100],ch2[100];
int i,j,k,a,b,count=0;
printf("input the string:");
scanf("%s",ch1);
if((fp=fopen("F:\\myfile.txt","w"))==NULL)
{
printf("Can not open the file!\n");
exit(0);
}
j=strlen(ch1);
fwrite(ch1,j,1,fp);
fclose(fp);
printf("input the baby string:");
scanf("%s",ch2);
k=strlen(ch2);
if(k>j)
{
printf("input again:");
scanf("%s",ch2);
}
for(i=0;i<j-k+1;i++)
{
a=i;
b=0;
while((ch1[a]==ch2[b]) && a<=k+i-1)
{
a++;
b++;
}
if(a-1==k+i-1)
count++;
}

printf("%d\n",count);
}