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  • CodeForces

    Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u

     Status

    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

    Input
    4 3 3
    2 3 1 1
    Output
    2
    Input
    1 1 1
    2
    Output
    0
    Input
    11 4 2
    2 2 0 7 3 2 2 4 9 1 4
    Output
    6

    Source

    模拟题
    #include <iostream>
    #include <algorithm>
    #include <string.h>
    #include <stdlib.h>
    #include <math.h>
    #include <stdio.h>
    using namespace std;
    int a[200050];
    int main()
    {
        int i,j,n,t,c,flag=0,ans=0;
        scanf("%d%d%d",&n,&t,&c);
        for(i=0;i<n;i++)
        cin>>a[i];
        for(i=0;i<n;i++)
        {
            if (a[i]>t)
            {
                if(flag>=c)
                ans=ans+flag-c+1;
                flag=0;
            }
            else if(a[i]<=t)
            flag=flag+1;
        }
        if(flag>=c)
        ans=ans+flag-c+1;
        printf("%d
    ",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Ritchie/p/5425275.html
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