Algorithm Training 2 recurses represents a time limit: 1.0s memory limit: 512.0MBThe 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 recursive side output this question is quite simple, but the original author is not me, but with me to practice the classmate, because, we two of the code compared to his better than mine?
#include <stdio.h>voidEintN) { intCur,i; for(i=0, cur=1; cur<=n;cur<<=1, i++); printf ("2"); for(;cur>0; cur>>=1, i--) { if(cur&N) {if(cur>2) {printf ("("); E (i); printf (")"); } if(cur==2) ; if(cur==1) printf ("(0)"); if((n-=cur)! =0) printf ("+2"); } }}intMain () {intA; scanf ("%d",&a); E (a); return 0;}
Blue Bridge Cup algorithm training 2 of recurses expression
