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
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Class:
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PrettyPrintingProduct
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Method:
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prettyPrint
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Parameters:
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int, int
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Returns:
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String
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Method signature:
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String prettyPrint(int A, int B)
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(be sure your method is public)
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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)
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Returns: "36288 * 10^2"
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1 * 2 * ... * 10 = 3628800 = 36288 * 10^2
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1)
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Returns: "7 * 10^0"
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The product of all numbers between 7 and 7,
inclusive, is obviously 7.
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2)
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Returns: "2038974024 * 10^0"
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For this input D has 10 digits.
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3)
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Returns: "28952...24024 * 10^0"
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For this input D has 11 digits. Note that we
output three dots even if just one digit is missing in the output.
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4)
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Returns: "2923450236 * 10^1"
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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.
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5)
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Returns: "14806...28928 * 10^1163"
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6)
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Returns: "12164...08832 * 10^3"
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Note that the last five digits of D can start with
a zero.
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7)
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Returns: "32382...26624 * 10^4"
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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