查找 "Cullen' 号的 C 代码”

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需要编写一段C代码，要求用户输入一个数字，该代码将检查该数字是否是“卡伦数”。

只要能通过“2^n * n + 1”计算出来的数字就是卡伦数。

卡伦数示例:

3=2^1 * 1 + 1
9=2^2 * 2 + 1
25=2^3 * 3 + 1

这是我正在编写的代码，有什么帮助吗？

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

int main(void)

{

    int num, brojP, potency = 0, numRepeats = 0, endResult=0, isCullen;

    printf("Unesite broj");
    scanf("%d", &num);

    do
    {

        potency = potency + 1; // initializing "potency" and at the same time making it one number larger at each repeat of the loop
        do
        {
            brojP = 2*potency;
            numRepeats = numRepeats + 1;
        } while (numRepeats < potency); // this entire loop is used for "2^n" part

        endResult = brojP * potency + 1; // calculate the "2^n * n + 1" 
        numRepeats = 0;

        if (endResult == num)
        {
            isCullen = 1;
            break;
        }


    } while (endResult < num);

    if (isCullen == 1)
        printf("Number inputted is Cullen's number\n");
    else
        printf("Number inputted isn't Cullen't number\n");

    return 0;


}

Answer 1

这个循环是错误的:

    do
    {
        brojP = 2*potency;
        numRepeats = numRepeats + 1;
    } while (numRepeats < potency); // this entire loop is used for "2^n" part

您每次都需要将上一次迭代的结果乘以 2，但实际上是将效力乘以 2。由于效力不会改变，因此您只是一遍又一遍地执行相同的任务。这样做:

    brojP = 1;
    for (numRepeats = 0; numRepeats < potency; numRepeats++) {
        brojP *= 2;
    }