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.
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.
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.
6 5
1 3 8 14 19 20
3
6 1
1 3 5 7 9 10
2
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.
题目链接：http://codeforces.com/problemset/problem/716/A
题目大意：
给你一个N(N<=100000)个字母敲击的时间a[i](a[i]<=109)，如果在M时间内没有敲击那么屏幕就清零，否则屏幕上就多一个字母，问最后屏幕剩下几个字母。
打下下一个字母的时候，如果和之前字母打下的时间不超过k的话，则保留前面的继续打，如果超过了，则前面的字母全部消失，只留下这一个字母。
下面给出AC代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 int a[100005]; 6 int n,t; 7 while(scanf("%d%d",&n,&t)!=EOF) 8 { 9 for(int i=0;i<n;i++) 10 scanf("%d",&a[i]); 11 int ans=1; 12 for(int i=0;i<n-1;i++) 13 { 14 if(a[i+1]-a[i]<=t) 15 ans++; 16 else ans=1; 17 } 18 printf("%d\n",ans); 19 } 20 return 0; 21 }
