Financial application- computing future tuition

标签（空格分隔）： 程序设计实验 c++

本人学院

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Description

Suppose that the tuition for Zhongshan Universtiy is 10000 this year and tuition increases 5% every year. Write a program that reads an integer n(n>0) and uses a loop to compute the cost of the tuition’s worth of the nth year.

If the answer is not a integer, cut off the digits.(e.g. 17.88 -> 17)

If the given n is more than 100, n = n % 100.

Input
An integer n (n>0).

Output
The total cost of four years worth of tuition starting n years from now.

You should set the precision to 2 and fixed.

Sample Input
10Sample Output
16288

Hint:
use cin & cout

读题

my answer

#include<iostream>using namespace std;int main() {    int n;    cin >> n;    n %= 100;    double tuition = 10000;    while (n--) {        tuition *= 1.05;    }    int ans = static_cast<int>(tuition);    cout << ans << endl;    return 0;}

the standard answer

#include<iostream>using std::cin;using std::cout;using std::endl;int main() {  double a = 10000;  int b;  cin >> b;  if (b > 100)    b = b % 100;  while (b--) {    a *= 1.05;  }  b = a;  cout << b << endl;  return 0;}