本文主要是介绍NUC1373 Bank Interest【水题】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Bank Interest
时间限制: 1000ms 内存限制: 65535KB
通过次数: 1总提交次数: 1
问题描述
Farmer John made a profit last year! He would like to invest it well but wonders how much money he will make. He knows the interest rate R (an integer between 0 and 20) that is compounded annually at his bank. He has an integer amount of money M in the range 100..1,000,000. He knows how many years Y (range: 0..400) he intends to invest the money in the bank. Help him learn how much money he will have in the future by compounding the interest for each year he saves. Print an integer answer without rounding. Answers for the test data are guaranteed to fit into a signed 32 bit integer.
输入描述
* Line 1: Three space-separated integers: R, M, and Y
输出描述
* Line 1: A single integer that is the number of dollars FJ will have after Y years.
样例输入
5 5000 4
样例输出
6077
来源
USACO 2004 November Gold
提示
INPUT DETAILS:
5% annual interest, 5000 money, 4 years
OUTPUT DETAILS:
Year 1: 1.05 * 5000 = 5250
Year 2: 1.05 * 5250 = 5512.5
Year 3: 1.05 * 5512.50 = 5788.125
Year 4: 1.05 * 5788.125 = 6077.53125
The integer part of 6077.53125 is 6077.
问题分析：（略）
这个问题和《POJ2390 Bank Interest【水题】》是同一个问题，代码直接用就AC了。
程序说明：参见参考链接。
参考链接：POJ2390 Bank Interest【水题】
题记：程序做多了，不定哪天遇见似曾相识的。AC的C++程序如下：
/* POJ2390 Bank Interest */#include <iostream>
#include <cmath>using namespace std;int main()
{int r, m, y;cin >> r >> m >> y;cout << (int)(m * pow(1 + r / 100.0, y)) << endl;return 0;
}
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