Power representation of Xidianoj 1024 2

Title Description

Any positive integer can be represented by a power of 2.
Example: 137=2^7+2^3+2^0
At the same time, it is agreed that the square is denoted by parentheses, that is, a^b can be represented as a (b)
The 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 1315 can be represented as: 2 (2 (+ + (0)) +2) +2 (2 (+ + (0)) +2 (2 (2) +2 (0)) +2+2 (0)

Input

Multiple groups, one positive integer per line (0<n<=20000)

Output

Each group of rows, conforming to the 0,2 representation of n (cannot have spaces in the representation)

--body recursion solves the problem of first turning a number into a binary sub-system to solve the power of 1 power times greater than 2 recursion that power is 2 or 0

#include <iostream>#include<cstdio>#include<cstring>using namespacestd;#defineSIZE 20intSolve (intN) {    intk[ -]; intTotal =0, temp =N;  while(Temp >1) { Total++; K[total]= temp%2; Temp/=2; }    if(temp = =1) { Total++; K[total] =1; }    intLast ; inti;  for(i=1; i<=total;i++){        if(K[i] = =1) { last=i;  Break; }    }     for(i=total;i>=1; i--){        if(K[i] = =0)Continue; intnow = i1; if(now = =1) {printf ("2"); }        Else{printf ("2 ("); if(Now >=3) Solve (now); Elseprintf"%d", now); printf (")"); }        if(I! =Last ) {printf ("+"); }    }}intMain () {intN;  while(SCANF ("%d", &n)! =EOF)    {Solve (n); }    return 0;}

Power representation of Xidianoj 1024 2