ACM-HDoj暑假竞赛（7）-1009解题报告(

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本文介绍了一个简单的数学问题：根据初始存款、年利率及目标金额，计算达到目标所需的年数。通过使用对数公式解决了该问题，并给出了C语言实现代码。

新浪博客发表时间 -- 2009-07-27 16:38:21)

题目：

Suppose you open a savings account with a certain initial balance. You will not make any withdrawals or further deposits for a number of years. The bank will compound your balance (add the annual interest) once a year, on the anniversary of the opening of the account. Your goal is to achieve a certain target amount in your savings account.

In how may years will the target amount be achieved?

Input：

The input file will contain data for one or more test cases, one test case per line. Each line will contain three numbers: the initial balance, the annual interest rate (as a percentage of the balance), and the target amount, separated by blank spaces. These will be positive numbers; they may or may not contain a decimal point.
The target amount will be greater than the initial balance. The input is terminated by end-of-file

Outp：

For each line of input, your program will produce exactly one line of output: This line will contain one positive integer value: the number of years required to achieve the target amount.

数据......

Sample Input：

200.00 6.5 300
500 4 1000.00

Sample Output：

算法：

虽然说英文不太懂，但一看到这个题目就大致猜到题意了，输入的数据是本金，利率，和最终所得的金额，求的是经过多少年能达到总金额，也就是第三个数.这应该是道很简单的初中数学题，幸好，还记得那个公式，一想到这就有点兴奋了，赶紧奋笔疾书...额，想想...P=(1+r%)^N*S...

P为总金额，r为每年的利润率，S为本金..有这公式就好办了...

代码：

     #include "stdio.h"
   #include "math.h"
int main()
{
int year;
float initial,rate,value,t,rate1;
while (scanf("%f %f %f",&initial,&rate,&value)!=EOF)
{
  year=0;
  rate1=rate/100;
  t=value/initial;
  year=log10(t)/log10(1+rate1);
  printf("%d\n",year+1);
}
}