[POJ1004]Financial Management
试题描述
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
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
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.
输入示例
100.00
489.12
12454.12
1234.10
823.05
109.20
5.27
1542.25
839.18
83.99
1295.01
1.75
输出示例
$1581.42
数据规模及约定
数字个数 = 12，结果和输入都在 double 范围内
题解
求 12 个数的平均数。
woc POJ 的 G++ 有毒！double 占位符是 "%.2f" 而不是 "%.2lf"，所以只能用 “C++”！！！
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std; int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
} int main() {
double sum = 0.0;
for(int i = 1; i <= 12; i++) {
double tmp;
scanf("%lf", &tmp);
sum += tmp;
} printf("$%.2lf\n", sum / 12.0); return 0;
}