SRM_420_div2_1000pt

SRM_420_div2_1000pt

Problem Statement

    

You will be given two positive ints A and B.

Let C be the product of all integers between A and B, inclusive.

The number C has a unique representation of the form C = D * 10^E, where D and E are non-negative integers and the last digit of D is non-zero.

Write a method that will return the value of C formatted as a String of the form "D * 10^E" (quotes for clarity only). Substitute the actual values of D and E into the output. If D has more than 10 digits, only output the first five and the last five digits of D, and separate them by three periods.

 

Definition

    

Class:	PrettyPrintingProduct
Method:	prettyPrint
Parameters:	int, int
Returns:	String
Method signature:	String prettyPrint(int A, int B)
(be sure your method is public)

     

Notes

-The purpose of the last constraint is to disallow inputs where precision problems could arise. 

Constraints

-A will be between 1 and 1,000,000, inclusive.-B will be between A and 1,000,000, inclusive.-If C has more than 10 digits, then the sixth most significant digit of C will be neither 0 nor 9. 

Examples

0)    

1 10

Returns: "36288 * 10^2"

1 * 2 * ... * 10 = 3628800 = 36288 * 10^2

1)    

7 7

Returns: "7 * 10^0"

The product of all numbers between 7 and 7, inclusive, is obviously 7.

2)    

211 214

Returns: "2038974024 * 10^0"

For this input D has 10 digits.

3)    

411 414

Returns: "28952...24024 * 10^0"

For this input D has 11 digits. Note that we output three dots even if just one digit is missing in the output.

4)    

412 415

Returns: "2923450236 * 10^1"

The actual value of C is larger than in the previous example. However, C ends in a zero and therefore D only has 10 digits.

5)    

47 4700

Returns: "14806...28928 * 10^1163"

6)    

1 19

Returns: "12164...08832 * 10^3"

Note that the last five digits of D can start with a zero.

7)    

13 25

Returns: "32382...26624 * 10^4"

 1 
 2  #include  < cmath >
 3  #include  < ctime >
 4  #include  < iostream >
 5  #include  < string >
 6  #include  < vector >
 7  using   namespace  std;
 8 
 9  typedef  long   long  i64;
10 
11  class  PrettyPrintingProduct {
12  public :
13      string  prettyPrint(  int  A,  int  B );
14  };
15  bool  check( int  A, int  B,i64 &  D,i64 &  E){
16      D = 1 ;E = 0 ;
17       for ( int  i = A;i <= B;i ++ ){
18          D *= i;
19           while (D % 10LL == 0LL) {
20              D /= 10LL;
21              E ++ ;
22          }
23           if (D > 10000000000LL)  return   false ;
24      }
25       return   true ;
26  }
27 
28  string  PrettyPrintingProduct::prettyPrint( int  A, int  B){
29      i64 D,tens;
30       char  str[ 256 ];
31 
32       if (check(A,B,D,tens)){
33          cout << D << "   " << tens << endl;
34          sprintf(str, " %lld * 10^%lld " ,D,tens);
35           return  str;
36      }
37 
38      i64 a = 1 ,b = 1 ;
39      tens = 0 ;
40       for ( int  i = A;i <= B;i ++ ){
41          a *= i;
42          b *= i;
43           while (b % 10LL == 0LL) b /= 10LL,tens ++ ;
44          b %= 1000000000000LL;
45           while (a >= 1000000000000LL) a /= 10LL;
46      }
47       while (a >= 100000LL) a /= 10LL;
48      b %= 100000LL;
49      cout << a << "   " << b << "   " << tens << endl;
50      sprintf(str,  " %lld%05lld * 10^%lld " , a, b, tens);
51       return  str;
52  }
53