给定一个数字 n，我们必须使用异或运算来打印将数字制成 2^X-1 形式的步骤。
- 我们应该进行异或任意 2^M-1 的数字，其中 M 由您选择，在奇数步长。
- 在偶数步长，将数字增加 1
继续执行该步骤，直到n变为2^X-1，并打印所有步骤
示例
Input: 22
Output:
Step 1 : Xor with 15
Step 2: Increase by 1
Step 3 : Xor with 7
Step 4: Increase by 1
Step 5 : Xor with 1
Input:7
Output: No Steps to be performed
算法
int find_leftmost_unsetbit(int n)
START
STEP 1 : DECLARE AND ASSIGN ind = -1, i = 1
STEP 2 : LOOP WHILE n
IF !(n & 1) THEN,
ASSIGN ind WITH i
END IF
INCREMENT i BY 1
LEFT SHIFT n BY 1
END WHILe
STEP 3 : RETURN ind
STOP
void perform_steps(int n)
START
STEP 1 : DECLARE AND ASSIGN left = find_leftmost_unsetbit(n)
STEP 2 : IF left == -1 THEN,
PRINT "No Steps to be performed"
RETURN
END IF
STEP 3 : DECLARE AND ASSIGN step = 1
STEP 4 : LOOP WHILE find_leftmost_unsetbit(n) != -1
IF step % 2 == 0 THEN,
INCREMENT n BY 1
PRINT "Step n : Increase by 1"
ELSE
DECLARE AND ASSIGN m =
find_leftmost_unsetbit(n)
AND SET num = (pow(2, m) - 1)
SET n = n ^ num
PRINT "Step N : Xor with Num
END IF
INCREMENT step BY 1
END LOOP
STOP
示例
#include <stdio.h>
#include <math.h>
//To find the leftmost bit
int find_leftmost_unsetbit(int n){
int ind = -1;
int i = 1;
while (n) {
if (!(n & 1))
ind = i;
i++;
n >>= 1;
}
return ind;
}
void perform_steps(int n){
// Find the leftmost unset bit
int left = find_leftmost_unsetbit(n);
//If there is no bit
if (left == -1) {
printf("No Steps to be performed");
return;
}
// To count the number of steps
int step = 1;
// Iterate till number is in form of 2^x - 1
while (find_leftmost_unsetbit(n) != -1) {
// if the step is even then increase by 1
if (step % 2 == 0) {
n += 1;
printf("Step %d: Increase by 1
", step);
}
// if the step is odd then xor with 2^m-1
else {
// Finding the leftmost unset bit
int m = find_leftmost_unsetbit(n);
int num = (int)(pow(2, m) - 1);
n = n ^ num;
printf("Step %d : Xor with %d
", step, num);
}
// To increase the steps
step += 1;
}
}
int main(){
int n = 22;
perform_steps(n);
return 0;
}
输出
<
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