The beauty of algorithms [from cainiao to expert drills]: some small algorithms

1. Number of completions within 10000

/** Problem description: calculate the number of completions within 10000. * End number: the sum of a number equal to its different factors * For example, 6 = 1 + 2 + 3.8! = 1 + 2 + 4 * xtfggef 2012/5/16 */# include <iostream> # include <fstream> # include <vector> using namespace STD;/** train of thought: calculate the number less than 10000, put it in * a vector, and compare it with the given number. If it is less than *, print it. Another idea is to directly calculate the number less than the given number, so that you can calculate less. */INT main (INT argc, char * argv []) {// introduce the file operation ifstream CIN ("seed.txt"); int N; vector <int>; /** here we need to explain why the for * loop below is I = I + 2, which indicates that the complete number is only in the even number, the odd number does not have a complete number in a small range. * The specific range does not exist in the mathematics field. */For (INT I = 2; I <10000; I = I + 2) {int sum = 1; for (Int J = 2; j <= I/2; j ++) {// if the Division is 0, it is a factor, so add if (I % J = 0) sum = sum + J;} If (sum = I). push_back (I) ;}while (CIN >>n) {cout <n <":"; for (int K = 0; k <. size (); k ++) {if (a [k] <= N) cout <"" <A [k] ;}cout <Endl ;} return 0 ;}

Baidu Encyclopedia of the number of: http://baike.baidu.com/view/640632.htm

2. Calculate the three-digit symmetric Prime Number

/** Problem description: Find a three-digit symmetric Prime Number * For example, 101 is, 787 is also, 896 is not * is input yes, otherwise, no * xtfggef 2012/5/16 */# include <iostream> # include <cmath> using namespace STD; bool isprime (INT); int main (INT argc, char * argv []) is output. {int N; CIN> N; // core cout <(n> 100 & n <1000 & N/100 = n % 10 & isprime (n )? "Yes \ n": "No \ n"); Return 0;}/** determine whether it is a prime number */bool isprime (int n) {int sqr = SQRT (N * 1.0); For (INT I = 2; I <= sqr; I ++) {If (N % I = 0) return false;} return true ;}

3. Calculate the maximum common divisor of two numbers.

/** Calculate the maximum common divisor of two numbers * xtfggef 2012/5/16 */# include <stdio. h> int gcd (INT, INT); void main () {int A = 5; int B = 8; int c = gcd (A, B ); printf ("% d \ n", c);} int gcd (int A, int B) {While (! = B) {If (A> B) A = A-B; elseb = B-A;} return ;}

The minimum public multiples are similar. With the help of the minimum public multiples = x * Y/gcd (x, y); then OK.

Note: to make the calculation not out of the range, it is best to write it as: X/gcd (x, y) * y

4. Replace the modulo operation method

K = (J + rotdist) % N
It can be written as follows:
K = J + rotdist;
If (k> N)
K-= N;

5. Calculate the largest sum of adjacent subvectors (from the programming pearl)

/** Find out the largest of any adjacent sub-vectors and * xtfggef 2012/5/7 */# include <stdio. h> int max (int A, int B) {return (A> B )? A: B;} void main () {int A [10] = {31,-, 26,-, 58, 97,-93 }; int maxsofar = 0; int maxendinghere = 0; For (INT I = 0; I <10; I ++) {// in case of a negative number, assign 0 directly, this is equivalent to skipping // a large number. maxendinghere = max (maxendinghere + A [I], 0); maxsofar = max (maxsofar, maxendinghere );} printf ("% d \ n", maxsofar );}

6. If an integer array a is given and the number of elements is N, find the element that appears n/2 or more times in a and print it out.

Package COM. xtfggef. hashmap; import Java. util. hashmap; import Java. util. map; import Java. util. set; /*** print the n/2 or more elements in the array * use a hashmap to store array elements and the number of occurrences * @ author erqing **/public class hashmaptest {public static void main (string [] ARGs) {int [] A = {,}; Map <integer, integer> map = new hashmap <integer, integer> (); For (INT I = 0; I <. length; I ++) {If (map. containskey (A [I]) {int TMP = map. get (A [I]); TMP + = 1; map. put (A [I], TMP);} else {map. put (A [I], 1) ;}} set <integer> set = map. keyset (); For (integer S: Set) {If (map. get (s)> =. length/2) {system. out. println (s );}}}}

7. Java implementation of hdoj problem 1000

import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        while(in.hasNextInt()){                int a = in.nextInt();            int b = in.nextInt();            int c = sum(a,b);            System.out.println(c);        }    }    public static int sum(int a,int b){        return a+b;    }}

8. hdoj problem 1001 C ++ implementation

#include <iostream>using namespace std;int main(){    unsigned int num;     while (cin >> num)    {        cout << (1+num)*num/2 << endl << endl ;     }    return 0;}

9. Java implementation of hdoj problem 1002

import java.math.BigInteger;import java.util.Scanner;public class Main{    public static void main(String[] args)     {        Scanner scanf = new Scanner(System.in);        while(scanf.hasNext())        {            int n;            int i=1,x=0;            n=scanf.nextInt();            while(n--!=0)           {               if(x++!=0)                   System.out.println();               BigInteger a,b,sum;               a=scanf.nextBigInteger();               b=scanf.nextBigInteger();               sum=a.add(b);               System.out.println("Case"+" "+ i++ +":");               System.out.print(a+" "+"+"+" "+b+" "+"="+" ");               System.out.println(sum);           }        }    }}

10. transpose the JAVA Implementation Matrix

Package COM. xtfggef. algo;/*** transpose of the matrix ** @ author erqing **/public class Matrix {public static void main (string [] ARGs) {int array [] [] = {1, 2, 3}, {4, 5, 6}, {7, 8, 9 }}; int array2 [] [] = new int [3] [3]; system. out. println ("transpose front:"); For (INT I = 0; I <array. length; I ++) {for (Int J = 0; j <array [I]. length; j ++) {system. out. print (array [I] [J] + ""); array2 [J] [I] = array [I] [J];} system. out. println ();} system. out. println ("transposed:"); For (int K = 0; k <array2.length; k ++) {for (INT h = 0; H <array2 [K]. length; H ++) {system. out. print (array2 [k] [H] + "");} system. out. println ();}}}

11. Time Difference in Java computing

import java.text.ParseException;import java.text.SimpleDateFormat;import java.util.Date;public class Tiemp {    public static void main(String args[]){        SimpleDateFormat sdf = new SimpleDateFormat("yyyyMMdd");        String s1 = "20080808";        String s2 = "20080908";        try {            Date d1 = sdf.parse(s1);            Date d2 = sdf.parse(s2);            System.out.println((d2.getTime() -  d1.getTime())/(3600L*1000));         } catch (ParseException e) {            e.printStackTrace();        }    }}

12. Determine IP Legality

1. From the number of. values, there cannot be more than three

2. The number between each vertex must be 0 ~ Between 255

3. Each number except. Must be <9 and> 0

4. The first and last characters cannot be.

# Include <stdio. h> int is_valid_ip (const char * IP) {int section = 0; // The decimal value of each section is int dot = 0; // Several separators int last =-1; // The if (* IP! = '. ') {While (* IP) {If (* IP = '. ') {dot ++; If (DOT> 3) {return 0;} If (section >=0 & section <= 255) {Section = 0 ;} else {return 0 ;}} else if (* IP >='0' & * IP <= '9 ') {Section = Section * 10 + * IP-'0'; If (last = '0') {return 0 ;}} else {return 0 ;}last = * IP; IP ++;} If (section >=0 & section <= 255) {If (3 = dot) {Section = 0; printf ("ip address success! \ N "); Return 0 ;}} return 1 ;}else {printf (" format error! "); Return 0 ;}} int main () {is_valid_ip (" 23.249.22.123 "); Return 0 ;}

13. judge whether the string is a return message.

# Include <stdio. h> # include <stdlib. h>/* determine whether the user-input string is a return object * a return object refers to a string with the same reading and reverse reading. * For example, abccba is a return object, abcdab is not a echo */INT palindrome (const char * Str) {int length = strlen (STR); For (INT I = 0; I <= length/2; I ++) {If (STR [I]! = STR [length-i-1]) {return-1 ;}} return 1 ;}int main () {char s [100]; gets (s ); int result = palindrome (s); If (result = 1) {printf ("the string is a 文"");} else {printf ("the string is not a 文 ");}}

14. convert a string to an integer

#include<iostream>#include<string>#include<assert.h>using namespace std;int str_2_int(string str){assert(str.size()>0);int pos = 0;int sym = 1;if(str[pos] == '+')pos++;else if(str[pos] == '-'){pos++;sym=-1;}int num =0;while(pos<str.length()){assert(str[pos]>='0');assert(str[pos]<='9');num = num*10+(str[pos]-'0');assert(num>=0);pos++;}num*=sym;return num;}int main(){string str = "-1024";int num = str_2_int(str);cout << num << endl;return 0;}