计划仅偶数值输出减去1

Question

所以我无法弄清楚为什么发生这种情况。该程序要求用户针对输入整数，取决于它是否分别是偶数或奇数任加倍数字或三倍的数字。问题是，对于只有偶数整数循环，它为每个循环迭代减去第一位数字1。奇数整数的循环与偶数整数相同，只是数字增加了三倍，而不是加倍，并且工作得很好。有任何想法吗？

例：

输入：12
预期输出：1122
实际输出：1121

输入：13
预期输出：111333
实际输出：111333

//This program asks the user to enter an integer and depending on whether the integer is even or odd, doubles or triples 
//each digit in the integer respectively. It then reads out the result and asks the user if they would like to enter 
//another integer and run the program again. 
int main() 
{ 
    string restart; 
    int integer, remainder; 

    while (restart != "n" && restart !="N") 
    { 
     cout << "Enter an integer: "; 
     cin >> integer; 
     cout << endl; 

     //Creates variable "temp" and "mycount" and divides "integer" by 10 until remainder is 0 and counts number of 
     //steps to do this to count the number of significant digits of the integer 
     int temp = integer, mycount = 0; 
     while (temp != 0) 
     { 
      temp = temp/10; 
      mycount++; 
     } 

     //Computes if the integer is even or odd by determining the remainder: 0 remainder = even, Non-0 remainder = odd 
     //Creates two integer variables "exponent" and "sum" and sets them to 0 
     //Assigns variable "temp" to the integer value 
     remainder = integer % 2; 
     int exponent = 0; 
     int sum = 0; 
     temp = integer; 

     //If integer is even, doubles each significant digit 
     if (remainder == 0) 
     { 
      //Begins for loop which runs for the number of times equal to the number of significant digits stored in "mycount" 
      for (int i = mycount; i > 0; i--) 
      { 
       //Stores current significant digit in "digit" 
       //Removes current significant digit by dividing by 10 
       //Multiplies current significant digit by 11 and stores it in "timesTwo" 
       //Multiplies current significant digit by 10^exponent then adds it to other modified digits 
       //Adds 2 to current exponent to increase digit multiplier by 100 
       int digit = temp % 10; 
       temp = temp/10; 
       int timesTwo = digit * 11; 
       sum = (timesTwo * pow(10, exponent)) + sum; 
       exponent = exponent + 2; 
       cout << sum << endl; 
       cout << endl; 
      } 

      cout << "Number is even, doubling each digit in the integer ..." << endl; 
      cout << sum << endl; 
     } 

     //If integer is odd this runs the same as the above function except it triples the digit and adds 3 to the multiplier 
     else 
     { 
      for (int i = mycount; i > 0; i--) 
      { 
       int digit = temp % 10; 
       temp = temp/10; 
       int timesThree = digit * 111; 
       sum = (timesThree * pow(10, exponent)) + sum; 
       exponent = exponent + 3; 
       cout << sum << endl; 
      } 
      cout << "Number is odd, tripling each digit in the integer ..." << endl; 
      cout << "Result: " << sum << endl; 
     } 

     cout << "Would you like to enter another integer? (y/n): "; 
     cin >> restart; 
    } 

    return 0; 
} 

Answer 1

为了调试这个，我会在第一个循环中打印所有输入，而不仅仅是第e输出。唯一可疑的事情，我可以看到的是，pow()返回一个浮点数，这可能会在转换回一个整数得到四舍五入。

为什么不避免pow()和隐式转换干脆使用你乘了每轮不是整数的因素？

sum = (timesTwo * factor) + sum; 
factor += 100; 

顺便说一句：这不是真的有必要算数字，并有不同的循环，对于两种情况 - 可以简化程序的核心，像

bool even = (integer & 1) == 0; 
int digitFactor = even ? 11 : 111; 
int stepFactor = even ? 100 : 1000; 
int result = 0; 
while(integer != 0) { 
    int digit = integer % 10; 
    result += digit * digitFactor; 
    integer /= 10; 
    digitFactor *= stepFactor; 
}