余りの合計が奇数の場合、11を加えない限り11 *を加算します。

Question

新しい問題が発生しました。 このプログラムの目標は、ユーザーが始めたのと同じA、B、Cの値で終わることです。私のコードは、例えば984や985のように、わずか3桁の整数を除いてちょうど3桁ごとに機能します。 "AとBの新しい値"は、9と8の代わりに3と2のように9と8の倍数になります（これはすべきです）。

問題の開始位置をコメントしました。それは私の新しい試みであり、コードの最初の試みです。前もって感謝します。

#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

ここでは完全な回答はありません。あなたの努力が示されて以来、私はあなたに完全な解決策を与えています。

ソリューション

#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 

問題

このコードブロックどんなにん、それは関係なく、条件文の常に実行されるので、newPは常にP/2がどうなるかifブロック

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; 

で上記。私はここでは、すべての可能性を組み込むことによって徹底的if-else組み合わせを取得している

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; 
} 

にそれを変更しました。 printステートメントの1つだけが1回の実行で実行できるようになりました。 newQとnewRのブロックはまったく同じです。

提案

コメントで示唆したように、完全にif-elseのアイデアをつかむしようとします。いくつかの基本的なコードを実行して動作を観察し、コードを実行していくつかのIDEでデバッグしようとします。