Input an integer to determine whether it is 2 ^ n. If it is not, it will output the integer that is closest to 2 ^ n.

Input an integer to determine whether it is 2 ^ n. If yes, output

// Number. If not, the output is the integer closest to 2 ^ n.

Additional source code 1:

# Include <stdio. h> # include <stdlib. h> # include <math. h> int main () {int input; // input an integer inputint I, j on the keyboard; // I, j. The difference between input and 2 ^ N on the left and right will be stored in the future: int M, N; // Save the value on the left, 2 ^ nprintf nearest to inout ("enter an integer input ="); scanf ("% d", & input); If (input <= 0) {printf ("not 2 ^ N"); printf ("2 ^ n nearest to it is 1");} else {for (int K = 0; k <1000; k ++) // assume that N in 2 ^ n will not exceed 1000 {If (input = POW (2.0, k )) {printf ("% d is 2 ^ K", input); break; // jumps out for loop} else if (input> POW (2.0, k-1) & input <POW (2.0, k) {I = input-Pow (2.0, k-1); j = POW (2.0, k)-input; M = POW (2.0, k-1); n = POW (2.0, k); if (I <j) // The number closest to him is the number on his left {printf ("% d ", m); break;} else {printf ("% d", n); break ;}}} system ("pause"); // include <stdlib. h> return 0 ;}

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Additional source code 2:

// If the number is the power of Npower of 2, the number can be first modulo 2 to 0, then perform Division 2, and then modulo 2,
// Until the last number is 2. If you cannot do this, the integer is not the Npower of 2.

<Span style = "color: #000000;"> # include <stdio. h> # include <stdlib. h> int main () {int input; // input an integer input int COUNT = 1 on the keyboard; // The integer to be judged is the count power of 2 printf ("enter an integer input ="); scanf ("% d", & input); While (input) {If (2 = input) // divide by 2 in sequence. If there is only 2 left, it indicates the power of 2 {printf ("yes: it is % d power \ n ", count); break;} If (0 = input % 2) of 2) // if the remainder of the number and 2 is 0 {input = input/2; // take half of the remainder and continue the loop count ++; // idempotent addition 1} else {printf ("NO \ n"); // This loop class can determine the closest 2 ^ n to it and output the break ;}} return 0 ;}</span>

Input an integer to determine whether it is 2 ^ n. If it is not, it will output the integer that is closest to 2 ^ n.