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. InputThe 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. OutputPrint 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.
模拟。倒序循环即可。
#include<cstdio>
#include<cstring>
int a[100010];
int main()
{int i,j,k,m,n,x,y,z,p,q,c;scanf("%d%d",&n,&c);for (i=1;i<=n;i++)scanf("%d",&a[i]);for (i=n-1;i;i--)if (a[i+1]-a[i]>c)break;printf("%d\n",n-i);
}