Common Algorithms for C Programming

{ t=strdr;
strdr=strdr[pi-1];
strdr[pi-1]=t;
}
strdr[p]='';
return(strdr);
}
main()
{ int x, d;
scanf("%d%d", &x, &d);
printf("%sn", trdec(x,d));
}

　　11. General processing of strings

　　1. Simple encryption and decryption
The idea of ​​encryption is: Add (or subtract) an ordinal number K to each letter C, that is, replace it with the Kth letter after it, the transformation formula is: c=c+k
For example, the ordinal number k is 5, then A→F, a→f, B→?G... When the letter after the ordinal number is added exceeds Z or z, then c=c+k -26
For example: You are good→ Dtz fwj ltti
Decryption is the inverse process of encryption.
Subtract (or add) an ordinal number K from each letter C, that is, c=ck.
For example, the ordinal number k is 5, then Z→U, z→u, Y→T... When the letter after the ordinal number is added is less than A or a, then c=ck +26 The
following program is the encryption processing:
#include
char *jiami(char stri[])
{ int i=0;
char strp[50],ia;
while(stri!='')
{ if(stri>='A'&&stri<='Z')
{ ia=stri+5;
if (ia>'Z') ia-=26;
}
else if(stri>='a'&&stri<='z')
{ ia=stri+5;
if (ia>'z') ia-=26;
}
else ia=stri;
strp[i++]=ia;
}
strp='';
return(strp);
}
main()
{ char s[50];
gets(s);
printf("%sn", jiami(s));
}
2. Count the words in a text
Input a line of characters and count the words in it. The words are separated by spaces.
Algorithm idea:
(1) Starting from the left side of the text (string), take a character; let the logical quantity word represent whether the character is a character in a word, and set the initial value to 0.
(2) If the character is not a word separator such as "space", "comma", "semicolon" or "exclamation mark", then determine whether word is 1. If word is not 1, it indicates the beginning of a new word. Let the word number num = num +1, and let word = 1;
(3) If the character is a word separator such as "space", "comma", "semicolon" or "exclamation mark", it indicates that the character is not a character in a word, and let word = 0;
(4) Take the next character in turn, and repeat (2) (3) until the end of the text.
The following program segment counts the number of words contained in the string string
#include "stdio.h"
main()
{char c,string[80];
int i,num=0,word=0;
gets(string);
for(i=0;(c=string)!='';i++)
if(c==' ') word=0;
else if(word==0)
{ word=1;
num++;}
printf("There are %d word in the line.n",num);
}

　　12. Exhaustive method
　　
　　The basic idea of ​​exhaustive method (also known as "enumeration method") is to list all possible situations one by one and judge which one may be the solution that meets the requirements. This is a "method when there is no other way" and is the most "dumb" method. However, it often works for some problems that cannot be solved by analytical methods. Usually loops are used to deal with exhaustive problems.
Example: A 100-yuan RMB note is exchanged for 100 5-yuan, 1-yuan and 0.5-yuan notes of equal value. It is required that each note must be at least 1. What are the combinations?
main()
{ int i, j, k;
printf(" 5 yuan 1 yuan 5 cents n");
for(i=1; i<=20; i++)
for(j=1; j<=100-i; j++)
{ k=100-ij;
if(5*i+1*j+0.5*k==100)
printf(" %3d %3d %3dn", i, j, k);
}
}

　　13. Recursive algorithm
　　
　　Using its own structure to describe itself is called recursion.
　　
　　VB allows calling itself within the definition of a Sub procedure and a Function procedure, that is, recursive Sub procedure and recursive Function function. Recursive processing is generally implemented using a stack. Each time it calls itself, the current parameter is pushed onto the stack until the recursive end condition is reached; then the current parameter is popped off the stack until the stack is empty.
Recursive conditions: (1) The recursive end condition and the value at the end; (2) It can be expressed in a recursive form, and the recursion develops towards the end condition.
Example: Write a recursive function fac(n)=n!
int fac(int n)
{ if(n==1)
return(1);
else
return(n*fac(n-1));
}
main()
{ int n;
scanf("%d", &n);
printf("n!=%dn", fac(n));
}