Interview Questions: quality factor decomposition

Question: Calculate the prime factor decomposition of a number, for example, input 90, and output 2 *3*3*5.
 

Disintegration ideas:

Requires a prime factor decomposition of number N. First, obtain all the prime numbers within N, put them in the prime [] array, and then let Prime [I] Remove n. If division can be performed, this prime [I] is the prime factor of N, otherwise it is not.

The prime [] method is to find the prime number by screening.

CodeImplementation

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# Include <iostream> # Include <Stdlib. h> Using   Namespace  STD;  Bool Isprime ( Int Number );//  Determine whether the number is a prime number  Void Printprime ( Int Number ); //  Search for all prime numbers within 0... number. No output is found.  Void Printarray ( Int Arry [], Int Len ); //  Print Array  Void Getprime2 ( Int Number, Int Prime []);//  Evaluate the prime number Array Using the isprime (INT number) Method  Void Getprime3 ( Int Number, Int Prime []); //  Implement isprime (INT number) directly in the Method)  //  Prime number obtained by screening  Int Getprime ( Int Number, Int  Prime []) {  Int Len = Number +1  ;  Int Count = 0  ;  Int  I, J;  Bool * Flag = New   Bool  [Len];  //  Initialize flag, all true;      For (I = 0 ; I <Len; I ++ ) {Flag [I] = True  ;}  For (I = 2 ; I <= SQRT (number); I ++) //  The prime factor of number must be less than SQRT (N)  {  If (Flag [I]) //  If I is a prime number, I * j is the Union number. These multiples are j = 2, 3, 4 ....  {  //  I indicates the prime factor, and J indicates a multiple.              For (J =2 ; I * j <= number; j ++) //  As long as I * j <= number, the flag is always assigned to the next value [];  {Flag [I * J] = 0  ;}}}  For (I = 2 ; I <= number; I ++ ){  If  (Flag [I]) {Prime [count ++] = I ;}} printarray (Prime, count );  Return Count ;}  //  Determine if it is a prime number  Bool Isprime ( Int  Number ){  Int  I;  Bool Flag = True  ;  For (I = 2 ; I <= SQRT (number); I ++) //  If number = 2, 3, The for loop cannot be executed, and the previous flag is returned, which is true. {  If (Number % I = 0  ) {Flag = False  ;  Break  ;}}  Return  Flag ;}  //  Print the prime number of 2... Number  Void Printprime ( Int  Number ){ Int Count = 0  ;  For ( Int I = 2 ; I <= number; I ++ ){  If  (Isprime (I) {count ++ ; Cout <I < Endl ;}}  //  Cout <count <Endl;  } //  Print Array  Void Printarray ( Int Arry [], Int  Len ){  For ( Int I = 0 ; I <Len; I ++ ) Cout <Arry [I] < "   "  ; Cout < Endl ;} //  Store the prime numbers within number in prime []  Void Getprime2 ( Int Number, Int  Prime []) {  Int Count = 0  ;  For ( Int I = 2 ; I <= number; I ++ ){  If (Isprime (I) {Prime [count ++] = I ;}} printarray (Prime, count );}  //  Store the prime numbers within number in prime []  Void Getprime3 ( Int Number, Int  Prime []) {  Int Count = 0  ;  For ( Int I = 2 ; I <= number; I ++ ){  Bool Flag = True  ;  For ( Int J = 2 ; J <= SQRT (I); j ++) //  Note that j <= SQRT (I) is used instead of j <= SQRT (number)  {  If (I % J = 0  ) {Flag =False  ;  Continue  ;}}  If  (FLAG) {Prime [count ++] = I ;}} printarray (Prime, count );}  Void  Main (){  Int Number = 90  ;  Int * Prime = New  Int  [Number];  //  Getprime2 (number, prime );  //  Store the prime number in the prime [] Array  //  Getprime3 (number, prime );  //  Store the prime number in the prime [] Array  //  Getprime (number );  //  Cout <isprime (number) <Endl;  // For bool type output, true = 1, false = 0  //  Printprime (number );      Int Len = getprime (number, prime ); //  Store the prime number in the prime [] Array      For ( Int I = 0 ; I <Len, number> = prime [I]; I ++) //  The number is smaller, and the prime [I] is larger. If the number is <Prime, it is not necessary to calculate  {  While (Number % prime [I] =0  ) {Number = Number/ Prime [I]; cout < Prime [I];  If (Number! = 1 ) //  If number = 1, it indicates that I is the last factor, so it does not need to be output * Cout < "  *  "  ;} Delete [] Prime; System (  " Pause  "  );}