POJ 2163 Easy Trading（我的水题之路——数组阶段n m的平均值比较）

Easy Trading

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1457		Accepted: 657		Special Judge

Description

Frank is a professional stock trader for Advanced Commercial Markets Limited (ACM Ltd). He likes "easy trading" -- using a straightforward strategy to decide when to buy stock and when to sell it. Frank has a database of historical stock prices for each day. He uses two integer numbers m and n (1 <= m < n <= 100) as parameters of his trading strategy. Every day he computes two numbers: P(m) -- an average stock price for the previous m days, and P(n) — an average stock price for the previous n days. P(m) > P(n) is an indicator of the upward trend (traders call it bullish trend), and P(m) < P(n) is an indicator of the downward trend (traders call it bearish trend). In practice the values for P(m) and P(n) are never equal. 
When a trend reverses from bearish to bullish it is a signal for Frank to buy stock. When a trend reverses from bullish to bearish it is a signal to sell. 

Frank has different values for m and n in mind and he wants to backtest them using historical prices. He takes a set of k (n < k <= 10 000) historical prices pi (0 < pi < 100 for 1 <= i <= k). For each i (n <= i <= k) he computes Pi(m) and Pi(n) — an arithmetic average of p _i-m+1 . . . p _i and p _i-n+1 . . . p _i respectively. Backtesting generates trading signals according to the following rules. 

• If Pi(m) > Pi(n) there is a bullish trend for day i and a "BUY ON DAY i" signal is generated if i = n or there was a bearish trend on day i - 1. 
  
• If Pi(m) < Pi(n) there is a bearish tread for day i and a "SELL ON DAY i" signal is generated if i = n or there was a bullish trend on day i - 1.

Your task is to write a program that backtests a specified strategy for Frank -- you shall print a signal for the first tested day (day n) followed by the signals in increasing day numbers. 

Input

The first line of the input contains three integer numbers m, n, and k. It is followed by k lines with stock prices for days 1 to k. Each stock price pi is specified with two digits after decimal point. Prices in the input file are such that Pi(m) != Pi(n) for all i (n <= i <= k).

Output

Write to the output a list of signals -- one signal on a line, as described in the problem statement.

Sample Input

3 5 17
8.45
9.10
9.40
10.15
10.40
11.08
11.52
12.12
12.51
12.15
11.90
11.25
11.73
10.77
10.80
10.01
9.14

Sample Output

BUY ON DAY 5
SELL ON DAY 12

Source

Northeastern Europe 2004

一个数组有k个元素，分别从第m个元素和第n个元素开始比较该元素之前的m、n个元素的平均值大小为p(m)、p(n)。如果p(m)>p(n)则当前的趋势为bullish tread，若i=n或i-1元素时为bearish tread，则输出"BUY ON DAY i"; 如果p(m)<p(n)则当前的趋势为bearish tread，若i=n或i-1元素时为bullish tread，则输出"SELL ON DAY i"。实时取值，比较后输出。分别用flag标记当前状态，-1为初始状态，0为bearish trend，1为bullish trend。

代码（1AC）：

#include <cstdio>
#include <cstdlib>
#include <cstring>

double arr[11000];

int main(void){
    int m, n, k;
    int i, j;
    double pm, pn;
    int flag; // 1 bullish trend ; 0 bearish trend

    while (scanf("%d%d%d", &m, &n, &k) != EOF){
        flag = -1;
        memset(arr, 0, sizeof(arr));
        pm = pn = 0;
        for (i = 0; i < k; i++){
            scanf("%lf", &arr[i]);
            if (i < m - 1){
                pm += arr[i];
            }
            else{
                pm -= arr[i - m];
                pm += arr[i];
            }
            if (i < n - 1){
                pn += arr[i];
            }
            else{
                pn -= arr[i - n];
                pn += arr[i];
                if (pm / m > pn / n && flag != 1){
                    printf("BUY ON DAY %d\n", i + 1);
                    flag = 1;
                }
                if (pm / m < pn / n && flag != 0){
                    printf("SELL ON DAY %d\n", i + 1);
                    flag = 0;
                }
            }
        }
    }
    return 0;
}