你的问题不够精确，尽管我觉得已经足够了。提供的信息要精确，没有必要显示其余的代码。
假设我们有char Ship[20]="ABCDEF";，如果你的编码像A=1，B=2一样简单，那么你只需要这样的代码：

char Ship[20]="ABCDEF";
int decoded=1;
for(int i=1; Ship[i]=!'/0'; ++i) {
    decoded = decoded * i
}
cout<<decoded;

这个循环会一直运行到字符串末尾的'\0'（空字符）。所以，你的代码（A=1，B=2，等等）代表阶乘，你就有了阶乘。
否则，您可以使用switch case或if语句来检查单个字符并进行适当解码。

char Ship[20]="ABCDEF";
int decoded=1;
for(int i=1; Ship[i]=!'/0'; ++i) {
    switch(Ship[i]){
        case 'A'  :   decoded = decoded * 1;
                      break;
        case 'B'  :   decoded = decoded *2;
                      break;
              //So on
        default   :   break;
    }
}
cout<<decoded;

两种情况下的输出：

720

将字母转换成数字（A=1; B=2......）”

string a{"ABC"};

    int a0 = a[0]; // 65
    int a1 = a[1]; // 66
    int a2 = a[2]; // 67
    .....

我想把A = 1，B = 2...

a0 = a0 - (65 - 1);
a1 = a1 - (65 - 1);
....

这个问题还不清楚，但我认为基本上是将char转换为int，它遵循A=1, B=2, ......, Z=26编码，并进行所需的处理，即乘以所有编码。
所以你可以这样做：

#include <iostream>
#include <string>

using namespace std;

int main(){
    string s;       //Input string 
    cout << "enter the string(CAPITALS ONLY) :";
    cin >> s;       //read the input string 
    int result = 1;
    for (auto &elem : s){          //process all the characters of s
        result *= elem - 'A' + 1;  //corresponding int value is multiplied to the result
    }

    cout << "the result is :" << result;
}

示例输出：

enter the string(CAPITALS ONLY) :AEF
the result is :30

（a+B）2=a2+b2+2ab：在cpp示例中

#include<iostream>

using namespace std;

//Declaring Function in scope

void firstFormula();

int main(void) //Main function
{
    cout << "Hello World!!" << endl; //sample test text printing

    firstFormula(); //executing function

    return 0;
}

//function implementation

void firstFormula(){

//initializing variables

    int a, b;

    cout << "Enter Value of A" << endl;

    cin >> a;//updating input values of a

    cout << "Enter Value of B" << endl;

    cin >> b;//updating input values of b

    int v1 = a + b; //(a+b)2

    int v2 = v1 * v1; //L.H.S

    cout << "Value of v1=" << v1 << endl << "V2=" << v2 << endl;

    cout << "Value of a=" << a << endl << "Value of B=" << b << endl;

    int v3 = a * a;

    int v4 = b * b;

    int v5 = 2 * a * b;

    int v6 = v3 + v4 + v5; //` R.H.S value after equation` `egfyufe`

    cout << "Output=" << "V3=" << v3 << endl << "V4=" << v4 << endl << "v5="
 << v5 << endl << "V6=" << v6 << endl;  

} //end