NYOJ 1242Interference Signal

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Interference Signal

时间限制：2000 ms | 内存限制：65535 KB

难度：1

描述

Dr.Kong’s laboratory monitor some interference signals. The interference signals can be digitized into a series of positive integer. May be, there are N integers a1,a2,…,an.

Dr.Kong wants to know the average strength of a contiguous interference signal block. the block must contain at least M integers.

Please help Dr.Kong to calculate the maximum average strength, given the constraint.

输入

The input contains K test cases. Each test case specifies:
* Line 1: Two space-separated integers, N and M.
* Lines2~line N+1: ai (i=1,2,…,N)
1 ≤ K≤ 8, 5 ≤ N≤ 2000, 1 ≤ M ≤ N, 0 ≤ ai ≤9999

输出

For each test case generate a single line containing a single integer that is 1000 times the maximal average value. Do not perform rounding.

样例输入

2 10 66 42103859415 210385 9

样例输出

65007333

来源

第八届河南省程序设计大赛

题目大意就是输入N个数然后从找出大于或等于M个连续的数的相加的最大平均值。。。其实题目比较水

输出的时候有坑，浪费了半个多小时。。。输出不能用printf("%.lf\n", max*1000); 要用%d输出

#include<stdio.h>int main() {  int k, n, m, a[2005], count, con;  scanf("%d", &k);  while(k--) {    double ave, max = -1.0;    scanf("%d%d", &n, &m);    for(int i = 0; i < n; i++)      scanf("%d", &a[i]);    count = m;    for(int t = 0; t <= n-m; t++) {      for(int i = 0; i <= n-m && i+count <= n; i++) {        double sum = 0.0;        for(int j = i; j < count+i; j++) {          sum += a[j];          //printf("%d ", a[j]);        }        ave = sum/count;        if(ave > max) max = ave;        //printf(" sum = %lf count = %d 平均值是:%lf\n", sum, count, ave);      }      count++;//控制求平均值数的个数     }    printf("%d\n", (int)(max*1000));  }}

这种题还是自己做比较好循环控制条件不自己去试不容易看懂

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