我认为相当难的问题...
只可以用stdio.h和string.h里的命令
我都快疯了，一点不会....
Universal Product Code Analysis
In this assignment you will write a C program to analyze a UPC
Barcode.
See: http://en.wikipedia.org/wiki/Universal_Product_Code for details.
There are a variety of UPC codes.
We will only analyze the 12 digit variety.
A barcode is printed on many items and is scanned in a shop for pricing and
inventory.
It consists of a sequence of black and white bars of different
widths.
The widths encode a sequence of 12 digits.
Here is a barcode for the UPC, 036000291452. As shown, there is a guard
at the start, a guard in the middle, and a guard at the end.
A black bar can be of width 1, 2, 3 or 4 and is represented by '1', '11',
'111' or '1111' Similarly a white bar is represented by '0', '00', '000' or
'0000'.
A valid barcode consists of the left guard, 101, a string of 42 '1's
or '0's representing the left 6 digits, a middle guard, 01010, a string of
42' 1's or '0's representing the right 6 digits, and the right guard, 101.
Each UPC digit is represented by 7 '1's or '0's.
Example: If 3 occurs in the
left 6 digits it is represented by 0111101, meaning a white bar of width 1,
a black bar of width 4, a white bar of width 1 and a black bar of width 1.
If 3 occurs in the right 6 digits it is represented by 1000010, (that is,
the "negative" of 0111101).
Your program will input a barcode from the file,
"barcode.dat". For the above barcode the file barcode.dat contains:
10100011010111101010111100011010001101000110
101010110110011101001100110101110010011101101100101
Your program will analyze this and convert it into the UPC code,
0 3 6 0 0 0 2 9 1 4 5 2.
The left values for all the ten digits are given by:
0 = 0001101, 1 = 0011001, 2 = 0010011, 3 = 0111101, 4 = 0100011,
5 = 0110001, 6 = 0101111, 7 = 0111011, 8 = 0110111, 9 = 0001011
These values can be represented by joining them into the string:
"00011010011001001001101111010100
01101100010101111011101101101110001011"
Note that the left value corresponding to digit d starts at position
7*d, and has length 7.
Similarly the right values is:
"1110010110011011011001000010101110
010011101010000100010010010001110100"
Your program will read a barcode consisting of 95 '0's or '1's, from the
file, barcode.dat,
and convert it into a 12 digit UPC. Assume that
there are no errors in the barcode.
Your program should use these functions:
main : read the barcode from barcode.dat, print its left and right parts,
call analyze, and print the resulting UPC digits.
void analyze(char barcode[ ], int upc[ ]): analyze the barcode given by
the first parameter and convert into 12 digits,
in the second parameter.
Calls convert.
int convert(char barcode[ ], int n, char code[ ]): convert the 7 barcode
characters starting at position n to a single UPC digit.
The first parameter,
barcode, should match part of the third parameter, code (which will either
be a left value string, or a right value string).
The position of the
match can be used to calculate the UPC digit.
Calls match.
int match (char a[ ], int aStart, char b[ ], int bStart, int n): match 'n'
characters of string a,
starting at index aStart with n characters of
string b starting at index bStart.
If successful, return 1 else
return 0.
The program should output the final UPC digits. Here is a sample output.
Barcode Analysis
Left part: 0001101 0111101 0111011 0100011 0101111 0101111
Right part: 1110010 1001000 1000010 1110010 1011100 1110100
UPC Code is: 0 3 7 4 6 6 0 8 3 0 4 9
[解决办法]
建议：
1。你要看懂英文的意思。如果看不懂就多读几遍。
2。涉及到打开读入的*。BAT文件，如果把里面的数据放到一个临时的存储区里面。然后从临时的存储区按照要求进行数据处理就可以了。
[解决办法]
他已经说得很明白了，连函数名和调用关系都指出来了，因该没什么难度了。
#define CODE_LEN 70 //0~9 barcode的长度char szBarcodeLeft[CODE_LEN]="00011010011001001001101111010100 01101100010101111011101101101110001011"; //左边6个barcode编码0～9char szBarcodeRigth[CODE_LEN]="1110010110011011011001000010101110 010011101010000100010010010001110100"; //右边6个barcode编码0～9#define SINGLE_LEN 7 //单个barcode长度void analyze(char barcode[],int upc[]){ int nUIndex=0; //数组upc的索引 int n=4; //由于完整的barcode前面有3个guard,所以从第四个开始 int nResult=-1; //convert返回的结果，小于零表示匹配失败，大于零则等于匹配的结果 for(;nUIndex<=5;n+=SINGLE_LEN) { nResult=convert(barcode,n,szBarcodeLeft); //转换左边6个 if(nResult>=0) { upc[nUIndex]=nResult; //将转换的结果装入upc数组 nUIndex++; } else { printf("no match barcode!") //他说假设barcode是正确的，所以一旦出现这种情况 break; //要么是读文件出错，要么是这段代码有问题。下同 } } for(n=51;nUIndex<=11;n+=SINGLE_LEN) //转换右边6个 { nResult=convert(barcode,n,szBarcodeRigth); if(nResult>=0) { upc[nUIndex]=nResult; nUIndex++; } else { printf("no match barcode!") break; } }}int convert(char barcode[],int n,char code[]){ int nIndex=0; for(int i=0;i<CODE_LEN-SINGLE_LEN;i+=SINGLE_LEN) //在数组中查找有无匹配 { if(match(barcode,n,code,i,SINGLE_LEN) //从0查起，直到查到匹配，这时的nIndex就是结果 return nIndex; nIndex++; } return -1; //没查到就返回无效值}int match(char a[],int aStart,char b[],int bStart,int n){ while(n>0) { if(a[aStart]!=b[bStart]) //每次连续匹配7个，有一个不同就返回0 return 0; aStart++; bStart++; n--; } return 1; //否则返回找到}
[解决办法]
我写了部分的代码，个人觉得已经把文章的要求理解透彻了
有点要注意就是，7个二进制的位应该对应的是2的7次方，但是这里就是0到9，有固定的转换编码，而且前6个的转换编码和后面6个是成反码的
我写了main和后面的一个转换的，再后面的两个很简单 了
// barcode.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <bitset.h>void analyze(char barcode[], int upc[]);int convert(char barcode[], int n, char code[]); int match(char a[], int aStart, char b[], int bStart, int n); int main(void) { FILE *stream; char Buffer[85]; int ch; // Open file to read line from: fopen("barcode.dat", "r" ); if( stream == NULL ) exit( 0 ); // 读取相应的左边的45个字符进Buffer[0--41], 右边的45个字符进rBuffer[42--83] ch = fgetc( stream ); for(int i=0; (i < 95 ) && ( feof( stream ) == 0 ); i++ ) { if ( (i > 2) && (i < 45) ) { Buffer[i - 3] = (char)ch; ch = fgetc( stream ); } if ( (i > 49) && (i < 92) ) { Buffer[i - 8] = (char)ch; ch = fgetc( stream ); } } Buffer[84] = '\0'; //打印相应的左右值 printf("Buffer = %s \n", Buffer); int upc[12] = {0}; //初始化每一个upc为0// 下面就是自己设计upc的转化了，和打印// 这里我省去了 fclose( stream ); return 0;}void analyze(char barcode[], int upc[]){ // 把barcode中每7个字符分析之后的值存入upc之中。 // analyze the barcode given by // the first parameter and convert into 12 digits. // 分别把其中的每7位按照顺序存入upc数组中就是了 static int j = 0; for (int i = 0; i < 84; i++) { upc[j] += barcode[i]; upc[j] = (upc[j] << 1); if ( !((i+1)%7) ) { j++; } } //施行转换，根据题目的要求，前面6个数字的0～9转化码与后面6个数字的转化码是不一样的， //其实仔细看一下就可以发现，其实转化码前6个与后6个是互为反码。 //The left values for all the ten digits are given by: //0 = 0001101, 1 = 0011001, 2 = 0010011, 3 = 0111101, 4 = 0100011, //5 = 0110001, 6 = 0101111, 7 = 0111011, 8 = 0110111, 9 = 0001011 // 右边取反码 for (int i = 0; i < 6; i++) { switch (upc[i]) { case 0001101： upc[i] = 0; break; case 0011001： upc[i] = 1; break; case 0010011： upc[i] = 2; break; case 0111101： upc[i] = 3; break; case 0100011： upc[i] = 4; break; case 0110001： upc[i] = 5; break; case 0101111： upc[i] = 6; break; case 0111011： upc[i] = 7; break; case 0110111： upc[i] = 8; break; case 0001011： upc[i] = 9; break; default: puts("不符合转换码规则，程序推出！"); exit(0); } } for (int i = 6; i < 12; i++) { switch (upc[i]) { case 0001101^1111111： //这里很简单，由于从第七个upc值到第十二个值的转换码对应着 //前6个的反码，我们让其值跟7个1异或便是对应的转换码 upc[i] = 0; break; case 0011001^1111111： upc[i] = 1; break; case 0010011^1111111： upc[i] = 2; break; case 0111101^1111111： upc[i] = 3; break; case 0100011^1111111： upc[i] = 4; break; case 0110001^1111111： upc[i] = 5; break; case 0101111^1111111： upc[i] = 6; break; case 0111011^1111111： upc[i] = 7; break; case 0110111^1111111： upc[i] = 8; break; case 0001011^1111111： upc[i] = 9; break; default: puts("不符合转换码规则，程序推出！"); exit(0); } } for (int i = 0; i < 12; i++) printf ("UPC Code is: %6d", upc[i]); puts("");}//先写这么多，觉得可以就给分吧//下面的很好实现了int convert(char barcode[], int n, char code[]){}int match(char a[], int aStart, char b[], int bStart, int n){}
[解决办法]
条形码
//开始标记：101
左边： 6个7位二进制 每一个7位二进制转换成以各数字（0--9） 根据左字典
中间标记01010
右边 6个7位二进制 每一个7位二进制转换成以各数字（0--9） 根据左字典
结束标记 101
//left字典
0 = 0001101, 1 = 0011001, 2 = 0010011, 3 = 0111101, 4 = 0100011,
5 = 0110001, 6 = 0101111, 7 = 0111011, 8 = 0110111, 9 = 0001011
右字典可以根据左字典得到，
右字典和左字典刚好互补，
如 左字典 0 = 0001101, 1 = 0011001
那么右字典 0 = 1110010 1 = 1100110
[解决办法]
#include "string.h"
#include "stdio.h"
enum NUMBER {ZERO, ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE};
struct
{
NUMBER index;
char code[8];
} left[10] =
{
ZERO, "0001101",
ONE, "0011001",
TWO, "0010011",
THREE, "0111101",
FOUR, "0100011",
FIVE, "0110001",
SIX, "0101111",
SEVEN, "0111011",
EIGHT, "0110111",
NINE, "0001011",
};
struct{
NUMBER index;
char code[8];
} right[10] =
{
ZERO, "1110010",
ONE, "1100110",
TWO, "1101100",
THREE, "1000010",
FOUR, "1011100",
FIVE, "1001110",
SIX, "1010000",
SEVEN, "1000100",
EIGHT, "1001000",
NINE, "1110100",
};
void analyze(char barcode[], int upc[]);
int convert(char barcode[], int n, char code[]);
int match(char a[], int aStart, char b[], int bStart, int n);
int main()
{
FILE* fp;
fp = fopen(".\\barcode.dat", "r");
while (NULL == fp)
{
fp = fopen(".\\barcode.dat", "r");
//Sleep(1);
}
char barcode[95] = {0};
int upc[12] = {0};
int ch;
int i=0;
ch = fgetc(fp);
while (i <95 && feof(fp) == 0)
{
barcode[i] = (char)ch;
ch = fgetc(fp);
i++;
}
//省打印相应左右值
analyze(barcode, upc);
for (i=0; i <12; i++)
{
printf("%d",upc[i]);
}
printf("\n");
printf("OK");
getchar();
return 1;
}
void analyze(char barcode[], int upc[])
{
int i,j,start=3;
for (i=0; i <6; i++)
{
for (j=0; j <10; j++)
{
if(convert(barcode, start, left[j].code))
{
upc[i] = j;
break;
}
}
start += 7;
}
start += 5;
for (; i <12; i++)
{
for (j=0; j <10; j++)
{
if(convert(barcode, start, right[j].code))
{
upc[i] = j;
break;
}
}
start += 7;
}
}
int convert(char barcode[], int n, char code[])
{
if (match(barcode, n, code, 0, 7))
{
return 1;
}
return 0;
}
int match(char a[], int aStart, char b[], int bStart, int n)
{
int i, nRet=0;
for (i=0; i <n; i++)
{
if (a[aStart] == b[bStart])
{
aStart++;
bStart++;
nRet++;
}
else
{
break;
}
}
return (nRet == n) ? 1 : 0;
}
注释暂时没写...