如果任何余数的和是奇数，则添加11 *除非*添加11导致

Question

我有新的问题了。这个程序的目标是最终得到与用户开始时相同的A、B、C值。我的代码几乎适用于每个3位数的整数，除了一些，比如984和985。“A和B的新值”是9和8的倍数，就像3和2一样，而不是9和8(就像它应该的那样)。

我已经对问题的起始点进行了评论。这是我的新尝试，也是我第一次尝试代码。提前谢谢。

#include <iostream>

using namespace std;
int main(int argc, char *argv[]) {

    //declare each variable
    int A,B,C;
    int ABC, BCA, CAB;
    int X,Y,Z; //remainders stored
    int P,Q,R; //sums of remainders

    //welcome
    cout<< "Welcome to Acelia's 3 digit reader \n"
        <<"\n";

    //prompt user
    cout << "Enter a number between 100 and 999: "; 
        cin >> ABC;

    //fits in range
    while ( (ABC < 100) || (ABC > 999) ) {
        cout << "Enter a valid number between 100 and 999: ";
        cin >> ABC;
    }

    cout << "Cool, you entered " << ABC 
         <<".\n"
         <<"\nIn the form of ABC...\n";

    //strip each digit of the number
    A = (ABC / 100);
    B = ((ABC/10) % 10);
    C = (ABC % 10);


    BCA = (B * 100 + C * 10 +A); //hundreds, tens, ones
    CAB = (C * 100 + A * 10 +B);
    cout <<"A is "<< A<< "\nB is " << B<<"\nC is " << C <<"\n"; //print individual #

    //print ABC, BCA, CAB
    cout <<"\nYour number in the form ABC is " << ABC
         <<"\nYour number in the form BCA is " << BCA
         <<"\nYour number in the form CAB is " << CAB
         << "\n\n";

    //store remainder of each value when divided by 11
    X = (ABC % 11); cout<< "The remainder of " << ABC <<" divided by 11 is " << X;
    Y = (BCA % 11); cout<< "\nThe remainder of " << BCA <<" divided by 11 is " << Y;
    Z = (CAB % 11); cout<< "\nThe remainder of " << CAB <<" divided by 11 is " << Z << "\n\n";

    //sums of each remainder
    P = (X + Y); cout<< "The sum of remainders from ABC and BCA is: " << P;
    Q = (Y + Z); cout<< "\nThe sum of remainders from BCA and CAB is: " << Q;
    R = (Z + X); cout<< "\nThe sum of remainders from CAB and ABC is: " << R <<"\n\n";

    int newP=0; //it would not execute properly w/o being initialized to 0
    int newQ=0;
    int newR=0;

    // Check sum remainder of X & Y 
    /*!!!!!!PROBLEM IS HERE!!!!!*/
    if (P % 2 == 1){

        if(P + 11 > 20) {
            newP = (P-11);

        }
        else {
            newP = (P+11);
        }

    }

    newP = (P/2);
    cout << "\nNew value of A is: " << newP; 

    //check sum remainder of Y & Z 
    /*ORIGNAL CODE */
    if ((Q % 2 == 1) && ((Q+11) > 20)){
            newQ = (Q-11);
            newQ = (newQ/2);
            cout<< ("\nNew value of B is: ") << newQ << "\n";
            if ((Q % 2 == 1) && ((Q + 11) < 20)) {
                newQ = (Q+11);
                newQ = (newQ /2);
                cout<< ("\nNew value of B is: ") << newQ << "\n";
            }
        }
        else {
            newQ = (Q/2);
            cout << "\nNew value of B is: " << newQ; 
        }


    //check sum remainder of Z + X
    if ((R % 2 == 1) && ((R+11) > 20)){
            newR = (R-11);
            newR = (newR/2);
            cout<< ("\nNew value of C is: ") << newR << "\n";
            if ((R % 2 == 1) && ((R + 11) < 20)) {
                newR = (R+11);
                newR = (newR /2);
                cout<< ("\nNew value of C is: ") << newR << "\n";
            }
        }
        else {
            newR = (R/2);
            cout << "\nNew value of C is: " << newR; 
        }

}//end of main

Answer 1

我们通常不会在这里提供完整的答案。既然你已经展示了你的努力，我将给你一个完整的解决方案。

解决方案

#include <iostream>

using namespace std;
int main(int argc, char *argv[]) {

    //declare each variable
    int A, B, C;
    int ABC, BCA, CAB;
    int X, Y, Z; //remainders stored
    int P, Q, R; //sums of remainders

    //welcome
    cout << "Welcome to Acelia's 3 digit reader \n" << "\n";

    //prompt user
    cout << "Enter a number between 100 and 999: ";
    cin >> ABC;

    //fits in range
    while ((ABC < 100) || (ABC > 999)) {
        cout << "Enter a valid number between 100 and 999: ";
        cin >> ABC;
    }

    cout << "Cool, you entered " << ABC << ".\n" << "\nIn the form of ABC...\n";

    //strip each digit of the number
    A = (ABC / 100);
    B = ((ABC / 10) % 10);
    C = (ABC % 10);

    BCA = (B * 100 + C * 10 + A); //hundreds, tens, ones
    CAB = (C * 100 + A * 10 + B);
    cout << "A is " << A << "\nB is " << B << "\nC is " << C << "\n"; //print individual #

    //print ABC, BCA, CAB
    cout << "\nYour number in the form ABC is " << ABC
            << "\nYour number in the form BCA is " << BCA
            << "\nYour number in the form CAB is " << CAB << "\n\n";

    //store remainder of each value when divided by 11
    X = (ABC % 11);
    cout << "The remainder of " << ABC << " divided by 11 is " << X;
    Y = (BCA % 11);
    cout << "\nThe remainder of " << BCA << " divided by 11 is " << Y;
    Z = (CAB % 11);
    cout << "\nThe remainder of " << CAB << " divided by 11 is " << Z << "\n\n";

    //sums of each remainder
    P = (X + Y);
    cout << "The sum of remainders from ABC and BCA is: " << P;
    Q = (Y + Z);
    cout << "\nThe sum of remainders from BCA and CAB is: " << Q;
    R = (Z + X);
    cout << "\nThe sum of remainders from CAB and ABC is: " << R << "\n\n";

    int newP = 0; //it would not execute properly w/o being initialized to 0
    int newQ = 0;
    int newR = 0;

    // Check sum remainder of X & Y
    /*!!!!!!PROBLEM IS HERE!!!!!*/
    if (P % 2 == 1) {

        if (P + 11 > 20) {
            newP = (P - 11);
            cout << "\nNew value of A is: " << newP;
        } else {
            newP = (P + 11);
            newP = (newP / 2);
            cout << "\nNew value of A is: " << newP;
        }

    } else {
        newP = (P / 2);
        cout << "\nNew value of A is: " << newP;
    }

    //check sum remainder of Y & Z
    /*ORIGNAL CODE */
    if (Q % 2 == 1) {
        if ((Q + 11) > 20) {
            newQ = (Q - 11);
            newQ = (newQ / 2);
            cout << ("\nNew value of B is: ") << newQ << "\n";
        } else {
            newQ = (Q + 11);
            newQ = (newQ / 2);
            cout << ("\nNew value of B is: ") << newQ << "\n";
        }
    } else {
        newQ = (Q / 2);
        cout << "\nNew value of B is: " << newQ;
    }

    //check sum remainder of Z + X
    if ((R % 2 == 1) && ((R + 11) > 20)) {
        if ((R + 11) > 20) {
            newR = (R - 11);
            newR = (newR / 2);
            cout << ("\nNew value of C is: ") << newR << "\n";
        } else {
            newR = (R + 11);
            newR = (newR / 2);
            cout << ("\nNew value of C is: ") << newR << "\n";
        }
    } else {
        newR = (R / 2);
        cout << "\nNew value of C is: " << newR;
    }

    return 0;

} //end of main

问题

在此代码块中

if (P % 2 == 1){

    if(P + 11 > 20) {
        newP = (P-11);

    }
    else {
        newP = (P+11);
    }

}

newP = (P/2);       //always executes regardless of the previous
                    //changes to the newP 
cout << "\nNew value of A is: " << newP;

无论if块做什么，newP都将始终为P/2，因为它将始终执行，而不管上面的条件是什么。我把它改成

if (P % 2 == 1) {

    if (P + 11 > 20) {
        newP = (P - 11);
        cout << "\nNew value of A is: " << newP;
    } else {
        newP = (P + 11);
        newP = (newP / 2);
        cout << "\nNew value of A is: " << newP;
    }

} else {
    newP = (P / 2);
    cout << "\nNew value of A is: " << newP;
}

在这里，我们通过整合所有的可能性得到了一个详尽的if-else组合。现在，在一次运行中只能执行一条print语句。newQ和newR的代码块完全相同。

建议

正如评论中所建议的，尝试完全抓住if-else的概念。运行一些基本代码来观察行为，并尝试通过单步执行代码在某些IDE中进行调试。