如何在沒有數學庫的情況下在 C++ 中創建復利公式？

Question

我一直在嘗試完成一項要求我運行復利公式的作業，但我只能使用 iomainip 和 iostream 庫。

do (cout << BaseMembershipFee * pow((ONE+(GoldRate/CompoundMonthly),(CompoundMonthly*years)) << years++) << endl;
    while (years < 11);

這適用於數學庫（我認為），但我不知道如何在沒有 pow 的情況下制作它。

變量：

const float BaseMembershipFee = 1200, GoldRate = 0.01, CompoundMonthly = 12, ONE = 1;
float years = 1;

Answer 1

在您的問題中，它指出年 = 1。基於 do while() 循環，我懷疑您的意思是顯示年 = 10。

下面的代碼有一個函數來計算每個時期的復利，在你的情況下是每個月。

然后，因為看起來你被要求在每年年底計算臨時金額，所以我們有第二個 for 循環。

#include<iostream>

float CompountInterestMultiplier(float ratePerPeriod, float periods) {
    float multiplier = 1.0;
    for (int i = 0; i < periods; i++) {
        multiplier *= (1.0 + ratePerPeriod);
    }
    return multiplier;
}

int main() {
    const float BaseMembershipFee = 1200;
    const float GoldRate = 0.01;  // represents annual rate at 1.0%
    const float periodsPerYear = 12.0;  // number of compounds per year.  Changed the name of this from compoundsMonthly
    const float ONE = 1.0;              // Did not use this
    const float years = 10.0;     // I think you have a typo 
                                  // and wanted this to be 10
    float ratePerPeriod = GoldRate / periodsPerYear;

    // If you want to print the value at the end of each year, 
    // then you would do this:
    float interimValue = BaseMembershipFee;
    for (int year = 0; year < 11; year++) {
        interimValue = BaseMembershipFee *
            CompountInterestMultiplier(ratePerPeriod, periodsPerYear * year);
        std::cout << "Value at end of year " << year << ": $" << interimValue << std::endl;
    }
}

Here is the output:

Value at end of year 0: $1200
Value at end of year 1: $1212.06
Value at end of year 2: $1224.23
Value at end of year 3: $1236.53
Value at end of year 4: $1248.95
Value at end of year 5: $1261.5
Value at end of year 6: $1274.17
Value at end of year 7: $1286.97
Value at end of year 8: $1299.9
Value at end of year 9: $1312.96
Value at end of year 10: $1326.15

Process finished with exit code 0