7-16 ZJW series of simple password (10)
, ZJW designed a program, the program's mission is to convert a string of phonetic alphabet to 6 digits (password),

ZJW can use any easy to remember the pinyin string (such as name, Zhong Jia Wei, just write: zhongjiawei) as input, the output 6 digits,

Program transformation process is as follows: the first step: put the pinyin string of a group of six fold, such as zhongjiawei into: zhongj iawei

The second step: put all the characters of vertical in the same location ascll code value addition, 6 Numbers, as the following example, concluded:

227 201 230 211 208 106

Step 3: for each digital processing, that is, each bit of the Numbers together, it is concluded that if it is not a single digit, so continue processing, until become a, for example: 227=& gt; 2 + 2 + 7=11=& gt; 1 + 1=2

After the above steps, the password is: 235417, this is the bank card password ZJW last

Input format:
The first line of input a T (0 & lt; T <100), said the number of sets of test data, a line, each group of data on behalf of a given pinyin string, length of not more than 20 to ensure are all lowercase English

The output format:
ZJW last bank card password, a set of data a line

Input the sample:
Given a set of input here, for example:

2
Zhongjiawei
Haohaoxuexi
The output sample:
Here is given the corresponding output, for example:

235417
875843

CodePudding user response:

Did you write this as a freshman? Difficulty is unlikely, but write a headache

 const int MAX=100; 
Const int LENTH=20; 
Char num [4] [6]={0}; 
Int j [6]={0}; 
Int m_deal [6]={0}; 
Int asccii () 
{
For (int j=0; J & lt; 6; J++) 
{
Int mo=0; 
For (int k=0; K & lt; 4. K++) 
{
Mo +=(int) num [k] [j]; 
} 
M [j]=mo; 
} 
return 0; 
} 
Int deal () 
{
Int a, b, c, d, e;//a 1000 10 
 \For (int j=0; J & lt; 6; J++) 
{
If (m [j] & gt; 
=100){
C=m/100 [j]; 
A=m [j] % 10; 
B=(m [j]/10) % 10; 
//three digits together the range is 0 to 27, so continue processing 
If (a + b + c & lt; 10) 
{
M_deal [j]=a + b + c; 
} 
The else//& gt;=10 
{
D=(a + b + c) % 10; 
E=(a + b + c)/10; 
If (d + e & lt; 10) 
{
M_deal [j]=d + e; 
} 
The else//the only one not to 10 is 19, 1 + 9=10, 1 + 0=1 
{
M_deal [j]=1; 
} 
} 
} 
The else//& lt; 100 
{
D=m [j] % 10; 
E=m [j]/10; 
If (d + e & lt; 10) 
{
M_deal [j]=d + e; 
} 
The else//the only one not to 10 is 19, 1 + 9=10, 1 + 0=1 
{
M_deal [j]=1; 
} 
} 
} 
return 0; 
} 
Int main (int arg c, _TCHAR * argv []) 
{
Int No; 
Char no [LENTH]={0}; 
Printf (" please enter the number of data: \ n "); 
scanf("%d",& No); 

for(int i=0; I & lt; No; I++) 
{
Char num_1 [6]={0};//after the superposition of string 
Printf (" please enter the first string: % d \ n ", I + 1); 
The scanf (" % s ", no); 
Int n=0; 
For (int j=0; J & lt; 4. J++) 
{
int m=0; 
While (m & lt; 6) 
{
Num [j] [m]=no [n]. 
M++; 
N++; 
If (n & gt;=20)//prevent an array 
break; 
} 
} 
//test 
//for (int I=0; I & lt; 4. I++) 
//{
//for (int j=0; J & lt; 6; J++) 
//printf (" % c ", num [I] [j]); 
//} 
Asccii (); 
The deal (); 
//test 
//for (int j=0; J & lt; 6; J++) 
//{
//printf (" % d \ n ", m_deal [j]); 
//} 
For (int j=0; J & lt; 6; J++) 
{
Printf (" % d ", m_deal [j]); 
} 
printf("\n"); 

//lowercase characters ASCII 97-122, so add up to the ASCII range is 97-122 * 4, namely three or two 

} 
return 0; 
} 

CodePudding user response:

How do you so many questions to test?

CodePudding user response:

Can I some simple:

 
#include  

Int main () 
{
For (int T=0, I=0, j, k [6]. ! T & amp; & ~ the scanf (" % d ", & amp; T); 
For (unsigned char [24] s={0}; I & lt; T & amp; & 24 ~ the scanf (" % s ", s); I++) 
{
For (j=0; J & lt; 6; J++) 
K [j] [j]=s + s + s + 6 [j] [j] + 12 + s + 18 [j]; 
For (j=0; J & lt; 6; J++) 
While (k [j] & gt; 9) k=k [j]/[j] [j] % 100/10 + 100 + k k [j] % 10; 
Printf (" % d % d % d % d % d % d \ n ", k [0], [1], k k [2], [3] k, k [4], [5] k); 
} 

return 0; 
}