nyoj 72-Financial Management (求和 ÷ 12.0)

72-Financial Management

内存限制:64MB 时间限制:3000ms 特判: No
通过数:7 提交数:12 难度:1

题目描述:

Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has decided that he needs to grab hold of his financial portfolio and solve his financing problems. The first step is to figure out what's been going on with his money. Larry has his bank account statements and wants to see how much money he has. Help Larry by writing a program to take his closing balance from each of the past twelve months and calculate his average account balance.

输入描述:

The input will be twelve lines. Each line will contain the closing balance of his bank account for a particular month. Each number will be positive and displayed to the penny. No dollar sign will be included. 

输出描述:

The output will be a single number, the average (mean) of the closing balances for the twelve months. It will be rounded to the nearest penny, preceded immediately by a dollar sign, and followed by the end-of-line. There will be no other spaces or characters in the output. 

样例输入:

复制

100.00
489.12
12454.12
1234.10
823.05
109.20
5.27
1542.25
839.18
83.99
1295.01
1.75

样例输出:

1581.42

题意：
　　将1~12月的money相加再除以12，得到的结果保留两位小数

C/C++代码实现（AC）;

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
const int MAXN = 310;
int A[MAXN];

int main()
{
    double a, cnt = 0.0;
    for(int i = 0; i < 12; ++ i)
    {
        scanf("%lf", &a);
        cnt += a;
    }
    printf("%.2lf", cnt / 12.0);
    return 0;
}