Codeforces Round #244 (Div. 2) B. Prison Transfer

B. Prison Transfer

The prison of your city has n prisoners. As the prison can't accommodate all of them, the city mayor has decided to transferc of the prisoners to a prison located in another city.

For this reason, he made the n prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.

Then, the mayor told you to choose the c prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,

• The chosen c prisoners has to form a contiguous segment of prisoners.
• Any of the chosen prisoner's crime level should not be greater then t. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.

Find the number of ways you can choose the c prisoners.

Input

The first line of input will contain three space separated integers n (1 ≤ n ≤ 2·10⁵), t (0 ≤ t ≤ 10⁹) and c (1 ≤ c ≤ n). The next line will contain n space separated integers, the i^th integer is the severityi^th prisoner's crime. The value of crime severities will be non-negative and will not exceed10⁹.

Output

Print a single integer — the number of ways you can choose the c prisoners.

Sample test(s)

Input

4 3 32 3 1 1

Output

2

Input

1 1 12

Output

0

Input

11 4 22 2 0 7 3 2 2 4 9 1 4

Output

6

求解连续序列的数目，并且其中的数有限制。

#include<iostream>//遍历一遍序列，当遇到不符合条件的数，判断一下是否满足ans累加，如果满足则tmp-m+1,代表当前可以划分为几个模块#include<cstring>#include<cstdio>//注意最后结束的时候也应当判断<span id="transmark"></span>一下#include<cmath>#include<algorithm>#define LL int#define inf 0x3f3f3f3fusing namespace std;LL a[200010];bool bj;int main(){    LL n,m,i,j,k,l,tmp;    while(~scanf("%d %d %d",&n,&l,&m))    {        for(i=0;i<n;i++)            scanf("%d",&a[i]);        LL ans=0;        tmp=0;        for(i=0;i<n;i++)        {            if(a[i]>l)            {                if(tmp>=m)                    ans+=tmp-m+1;                 tmp=0;            }            else            tmp++;        }        if(tmp>=m)        ans+=tmp-m+1;        printf("%d\n",ans);    }    return 0;}