The problem describes that any positive integer can be represented by a 2 binary, for example: 137 of the 2 binary is represented as 10001001.
The 2 notation is written in the form of the sum of the power of 2, and the second power is preceded by the following expression: 137=2^7+2^3+2^0
The Covenant power is now represented by parentheses, which means that a^b is represented as a (b)
At this point, 137 can be represented as: 2 (7) +2 (3) +2 (0)
Further: 7=2^2+2+2^0 (2^1 with 2)
3=2+2^0
So the last 137 can be represented as: 2 (2 (2) +2+2 (0)) +2 (+ 0) +2 (0)
Another example: 1315=2^10+2^8+2^5+2+1
So the last 1315 can be expressed as:
2 (2 (+ 0) +2) +2 (2 (+ + (0))) +2 (2 (2) +2 (0)) +2+2 (0) The input format positive integer (1<=n<=20000) output format conforms to the contract N of 0, 2 = (no spaces in the representation) sample input 137 sample Output 2 (2 (2) +2+2 (0)) +2 (0) +2 Example input 1315 sample output 2 (2 (+2 (0)) +2) 2 (0 (+ +2) 2 (2)) +2+2 (0) Prompt with recursive implementation will be relatively simple, can be output on the side of recursion
Answer:Before solving such problems, the first thing to think about is recursive implementation; In this case, we first need to find the similarity of this problem: in the topic of the hint: there is a binary way to represent this number, but we can not solve the problem by the conversion of the binary. So we solve them through tentative thinking. By a comparison of the current value with the N power of 2, a power of n is found that is exactly less than the current value of 2. Which means the problem is recursive.
Similarity ofIs:
Current value = 2 n Power +f (current value-2 n Power) just below the current value;The recursive
ExportRelatively easy to find: Export one: The number of the time: Output 2 (0); exit Two: The number is two when: Output 2; The following is the source code:
#include <iostream>
#include <cmath>
using namespace Std;
void Show (int b) {
Export Point Processing
if (b==1) {
cout<< "2 (0)";
return;//must be returned otherwise it will die loop
}
else if (b==2) {
cout<< "2";
return;
}
Similarity point Processing
for (int i=2;i<=15;i++) {//2 0, 1 power equals 1, 2, so find from 2
if (Pow (2,i) >b) {
if (i-1==1) {
Show (2);
}else{
cout<< "2 (";
Show (I-1);
cout<< ")";
}
if (B-pow (2,i-1) >0) {
cout<< "+";
Show (B-pow (2,i-1));
}
Break
}
}
}
int main () {
int b;
cin>>b;
Show (b);
return 0;
}
Recurses Representation of algorithm training 2 (Blue Bridge Cup C + + notation)
