South Bridge--recurses expression of algorithm training 2

Topic details

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)

Tips
Recursive implementation will be relatively simple, can be output on the side of recursion
Idea: 1, first convert the input 10 number into 2 binary number

2, the binary array bitwise processing, the original decimal number is represented as the corresponding binary representation (for example:137=2^7+2^3+2^0), note the Red flag content

3, judge whether the index can continue to be expressed as a binary system, is the title of the recursive

Note: 1, binary array from the back forward output is the original decimal binary form, so the processing time to start from the last.

2, ^ conversion to (), to increase the output control (), 2^1 to direct output 2, and for 2^0 and 2^2 respectively output 2 (0) and 2 (2).

3, for the control of the Plus, the last one after processing does not need to output a plus, from a one start behind all 0, the bit after processing does not need to output a plus.

Code:

#include <stdio.h>void solve (int n) {//Recursive exit//when exponent is 0 or 2 o'clock, direct output exponent if (n==0 | | n==2) {printf ("%d", n); return;} When the exponent is 1 o'clock (i.e. 2 (1) without output) if (n==1) {return;}/******* above is recursive exit ********/char Str[20];int i=0,j,len,q;//is converted to 2 binary, the resulting string is the original flashback arrangement While  (n) {str[i]=n%2+ ' 0 '; n/=2; i++;} len = i;/*while () is used to mark the count of 0, which is used to control the output of the + sign */q=0; while (str[q]== ' 0 ') {q++;} /  * for 2 binary string bitwise processing *////////variable J to indicate the current processing is the number of the output */for (i--, j=1;i>=0;i--, j + +) {if (str[i]!= ' 0 ') {/* If the currently processed character has a bit ( The bit is 2 (1), then the parentheses are omitted) */if ((len-j) ==1) {printf ("2");} Else{printf ("2 (");} /* Recursive processing */solve (LEN-J);/* If there is a bit after the currently processed character (the bit is 2 (1), the parentheses are omitted) */if ((len-j) ==1); else{printf (")");} /* The last one is currently processed, or the back of the current bit is all 0 without output +*/if ((len-j) ==0 | | i==q); else{printf ("+");}}} int main () {int n;scanf ("%d", &n),//1 handles if (n==1) {printf ("2 (0)") separately;} Else{solve (n);} return 0;}

South Bridge--recurses expression of algorithm training 2