codeforces --- Round #244 (Div. 2) B. Prison Transfer

思路摘自http://www.cnblogs.com/acmer-jsb/p/3707989.html

这题思路还是很巧妙的，遍历一遍，碰到超出限制的数再进行判断即可

#include <stdio.h>
#include <iostream>
using namespace std;
int main(){
    int n, t, c, i, j, k;
    int ans = 0, cnt = 0;
    cin >> n >> t >> c;
    for(i = 0; i < n; ++i){
        scanf("%d",&k);
        if(k > t){
            if(cnt >= c)    ans += cnt - c + 1;
            cnt = 0;
        }
        else
            cnt++;
    }
    if(cnt >= c)    ans += cnt - c + 1;
    cout << ans << endl;
    return 0;
}

接下来是我的过了初判最后TLE的CODE：

 #include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
const int MAXSIZE=50002;
const int INF=0x3fffff;
int main(){
    int i, j, k, n, m, tmp,t ,c;
    int pol,thi,ans;
    int start = 0, end;
    int array[222222];
    ans = 0;
    scanf("%d%d%d",&n,&t,&c);
    for(i = 0; i < n; i++)  scanf("%d",&array[i]);
    int flag = 0;
    for(end = start + c - 1; end < n; start++,end++){
        if(!flag){
            for(m = start; m <= end ; m++){
                if(array[m] > t)    break;
            }
            if(m == end + 1){
                ans++;
                flag = 1;
            }
        }
        else{
            if(array[end] <= t) ans++;
            else    flag = 0;
        }
    }
    printf("%d
",ans);
    return 0;
}