Codeforces-716A-Crazy Computer

最新推荐文章于 2021-07-28 15:30:12 发布

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最新推荐文章于 2021-07-28 15:30:12 发布

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原文链接：http://www.cnblogs.com/vocaloid01/p/9514223.html

本文介绍了一个有趣的编程问题，即在一个特殊的电脑上打字，如果超过一定时间未输入新的单词，屏幕上所有已输入的内容将被清除。文章通过一个示例详细解释了规则，并提供了一段C++代码来解决这个问题。

Time limit
2000 ms
Memory limit
262144 kB

ZS the Coder is coding on a crazy computer. If you don't type in a word for a c consecutive seconds, everything you typed disappear!

More formally, if you typed a word at second a and then the next word at second b, then if b - a ≤ c, just the new word is appended to other words on the screen. If b - a > c, then everything on the screen disappears and after that the word you have typed appears on the screen.

For example, if c = 5 and you typed words at seconds 1, 3, 8, 14, 19, 20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.

You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.

Input

The first line contains two integers n and c (1 ≤ n ≤ 100 000, 1 ≤ c ≤ 109) — the number of words ZS the Coder typed and the crazy computer delay respectively.

The next line contains n integers t1, t2, ..., tn (1 ≤ t1 < t2 < ... < tn ≤ 109), where ti denotes the second when ZS the Coder typed the i-th word.

Output

Print a single positive integer, the number of words that remain on the screen after all n words was typed, in other words, at the second tn.

Example
Input

6 5
1 3 8 14 19 20

Output

3

Input

6 1
1 3 5 7 9 10

Output

2

Note

The first sample is already explained in the problem statement.

For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;  

int data[100005];

int main()  
{  
    int N,C;
    cin>>N>>C;
    for(int i=0 ; i<N ; i++)scanf("%d",&data[i]);
    int sum = 1;
    int flag = data[0];
    for(int i=1 ; i<N ; i++)
    {
        if(data[i]-flag>C)sum = 0;
        flag = data[i];
        sum++;
    }
    cout<<sum<<endl;
    return 0;   
}