C++初学者,两道实验题目希望大神帮忙解答，多谢

[崖下人生]

Part I.
A person invests $1000.00 in a saving account yielding 5% interest. Assuming that all
interest is left on deposit in the account, calculate and print the amount of money in the
account at the end of each year for 10 years. Use the following formula for determining
these amounts: a=p(1+r)^n where p is the original amount of money invested (i.e., the
principle), r is the annual interest rate, n is the number of years and a is the amount of
deposit at the end of the nth year. Display the result in a table. Make sure the columns are
neatly aligned.

Part II.

Legend has it that, in 1626, Peter Minuit purchased Manhattan Island for $24.00 in barter.
Did he make a good investment? To answer this question, modify part I to begin with
p=24.00 and to calculate the amount of deposit a until this year (i.e., 2012). You just need
to print the value of a every other 50 years and 2012 (i.e., years 1676, 1726, …, 2012).
Try to use different interest rates from 5%, 6%, …, to 10%.

[vcf_reader]

That is too simple.

[Falleyes]

其实只要读懂题目就很简单的，小学数学题啊。 建议楼主用在线段落翻译，然后自己做吧。

[崖下人生]

引用 1 楼 vcf_reader 的回复:

That is too simple.

I know that it is so easy. However, It is different to do by using C++ for beginner.Therefore,I hope you can help me. Thank you very much

[崖下人生]

引用 3 楼 u013601512 的回复:

[quote=引用 1 楼 vcf_reader 的回复:] That is too simple.

I know that it is so easy. However, It is difficult to do by using C++ for beginner.Therefore,I hope you can help me. Thank you very much

[u012723504]

你先翻译一下再做，很简单的

[auq404]

#include <iostream> #include <math.h> using namespace std; double calc_money(double original_money,double rate,int year); int main() { int year = 10; double original_money= 1000; double rate = 0.05; double result = 0; result = calc_money(original_money,rate,year); return 0; } double calc_money(double original_money,double rate,int year) { double result = 0; result = original_money*pow(1+rate,year); return result; }

[Falleyes]

PartⅠ:

#include<iostream>
#include<cmath>
using namespace std;

const double p = 1000.00;
const double r = 0.05;

double cal(int n){
	return p*pow(1 + r, n);
}

int main(){
	for (int i = 0; i < 10; i++){
		cout << i + 1 << '\t';
		cout.setf(ios::fixed, ios::floatfield);
		cout.precision(2);
		cout<< cal(i + 1) << endl;
	}
}

Part Ⅱ:

#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;

const double p = 24.00;
const double r[] = { 0.05, 0.06, 0.07, 0.08, 0.09, 0.10 };

double cal(int n,int j){
	return pow(1 + r[j], n);
}

int main(){
	int start_year = 1626;
	double rate[6];
	for (int i = 0; i < 6; i++)
		rate[i] = cal(50, r[i]);
	for (int k = 0; k < 6; k++){
		cout << "年份\t利率"<<r[k] << endl;
		for (int i = 1; i < 8; i++){
			cout << start_year + i * 50 << '\t';
			cout.setf(ios::fixed, ios::floatfield);
			cout.precision(2);
			cout <<p*pow(rate[k], i) << '\t';
			cout << endl;
		}
		cout << "2012\t";
		cout.setf(ios::fixed, ios::floatfield);
		cout.precision(2);
		cout << p*pow(rate[k], 7)*cal(36, k) << endl;
		cout << endl;
	}
}

最后的结果你再检查检查，仅供参考

[崖下人生]

尽管语言有时很苍白，但是我非常诚恳的向楼上两位表示感谢