Description
The prison of your city has n prisoners. As the prison can‘t accommodate all of them, the city mayor has decided to transfer c 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 ith integer is the severity ith prisoner‘s crime. The value of crime severities will be non-negative and will not exceed 109.
Output
Print a single integer — the number of ways you can choose the c prisoners.
Sample Input
4 3 32 3 1 1
1 1 12
11 4 22 2 0 7 3 2 2 4 9 1 4
Sample Output
2 0 6
思路
题解:
试了一发暴力，果断超时，后来想了下，我们只需要记录下超过t的数，然后根据相邻两次超过t的数的位置求得此段不超过t的最长区间，看看是否符合题意，然后就可以计算出有多少种方法了。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int a[maxn], pos[maxn] = {0};
int main()
{
int n, t, c, p = 1, res = 0;
scanf("%d%d%d", &n, &t, &c);
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
if (a[i] > t) pos[p++] = i;
}
pos[p++] = n + 1;
if (pos[1] == 0) res += 1 + (n - c);
else
{
for (int i = 1; i < p; i++)
{
int len = pos[i] - pos[i - 1] - 1;
if (len >= c)
{
res += 1 + (len - c);
}
}
}
printf("%d\n", res);
return 0;
}