8758: The power of 2 is expressed, and the power of 8758 is expressed

8758: The power of 2 is expressed, and the power of 8758 is expressed
Power of 8758: 2

• View
• Submit
• Statistics
• Question

Total time limit:

1000 ms

 

Memory limit:

65536kB

Description

Any positive integer can be expressed by the power of 2. For example:

137 = 27 + 23 + 20

At the same time, the conventions are represented by parentheses, that is, AB can be expressed as a (B ). 137 can be expressed:

2 (7) + 2 (3) + 2 (0)

Further: 7 = 22 + 2 + 20 (expressed as 2 in 21)

3 = 2 + 20

So the last 137 can be expressed:

2 (2 (2) + 2 + 2 (0) + 2 (2 + 2 (0) + 2 (0)

Another example:

1315 = 210 + 28 + 25 + 2 + 1

So 1315 can be expressed:

2 (2 (2 + 2 (0) + 2) + 2 (2 (2 + 2 (0) + 2 (2 (2) + 2 (0 )) + 2 + 2 (0)

Input

A positive integer n (n ≤ 20000 ).

Output

A row. It is equal to 0, 2 of n (no space is allowed in the representation ).

Sample Input

137

Sample output

2(2(2)+2+2(0))+2(2+2(0))+2(0)

Source

Question 1 of NOIP1998 semi-finals popularity Group

1 # include <cstdio> 2 # include <cstdlib> 3 # include <cstring> 4 # include <cmath> 5 void work (int n) 6 {7 if (n = 1) 8 {9 printf ("2 (0)"); 10 return; 11} // initial judgment condition, if n is 1 or 2, 12 else if (n = 2) 13 {14 printf ("2"); // The 15return of 2; 16} 17 else 18 {19 int j = 1, I = 0; // j multiplied by 2 each time. If n is greater than, the decomposition ends, 20 // I is the current number (INDEX) 21 do 22 {23 j = 2 * j; 24 if (j> n) 25 {26 j/= 2; // roll back to when j <n at 27 if (I = 1) // this step is very important to determine whether to continue 2 () 28 printf ("2 "); 29 else 30 {3 1 printf ("2 ("); 32 work (I); 33 printf (")"); 34} 35 if (n-j! = 0) 36 {// if there is a remaining number after n decomposition, continue to decompose 37 printf ("+"); 38 work (n-j); 39} 40 return; 41} 42 else43 I ++; 44} while (1); 45} 46} 47 int main () 48 {49 int n; 50 scanf ("% d ", & n); 51 work (n); 52}Click here to view the answer