Codeforces #211 (Div. 2) B. Fence

B. Fence

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

There is a fence in front of Polycarpus's home. The fence consists of n planks of the same width which go one after another from left to right. The height
of the i-th plank is hi meters,
distinct planks can have distinct heights.


Fence for n = 7 and h = [1, 2, 6, 1, 1, 7, 1]

Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly k consecutive
planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such kconsecutive planks that the sum of their heights
is minimal possible.

Write the program that finds the indexes of k consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it
is in front of home (in other words, the fence isn't cyclic).

Input

The first line of the input contains integers n and k (1 ≤ n ≤ 1.5·105, 1 ≤ k ≤ n)
— the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers h1, h2, ..., hn (1 ≤ hi ≤ 100),
where hi is
the height of the i-th plank of the fence.

Output

Print such integer j that the sum of the heights of planks j, j + 1,
..., j + k - 1 is the minimum possible. If there are multiple such j's,
print any of them.

Sample test(s)

input

7 3
1 2 6 1 1 7 1

output

3

Note

In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

题目还是比较水的，就是求每几个数的和，然后取最小值。

刚开始用二次循环来做，超时了。

后来发现可以这样处理：

用sum[0]保存a[0]到a[k-1]项的和，那么sum[1]就可以表示为sum[0]-a[0]+a[k]，以此类推。

把时间复杂度降到了O(n)；

代码如下：

#include <stdio.h>
int main(void)
{
int n,k,i,j,a[150010],sum[150010],flag;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum[i]=0;
}
int min;
for(i=0;i<k;i++)
sum[0]+=a[i];
min=sum[0];
flag=0;
for(i=1;i<n-k+1;i++)
{
sum[i]=sum[i-1]-a[i-1]+a[k+i-1];
if(min>sum[i])
{
min=sum[i];
flag=i;
}
}
printf("%d\n",flag+1);
}
return 0;
}