NanoApe Loves Sequence Ⅱ(尺取法)

最新推荐文章于 2022-04-26 19:05:44 发布

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最新推荐文章于 2022-04-26 19:05:44 发布

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文章标签： java

原文链接：http://www.cnblogs.com/littlepear/p/5745200.html

本文提供NanoApeLovesSequenceⅡ题目解析及代码实现，该问题涉及寻找序列中满足特定条件的连续子序列数量，通过使用前缀和与滑动窗口的方法进行高效求解。

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题目链接：NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 339 Accepted Submission(s): 165

Problem Description

NanoApe, the Retired Dog, has returned back to prepare for for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with

Input

The first line of the input contains an integer

Output

For each test case, print a line with one integer, denoting the answer.

Sample Input

1 7 4 2 4 2 7 7 6 5 1

Sample Output

题解：

前缀和真是个好东西。前缀和可以是前n个的和，前n项的最小值，前n项的最大值。。。

这个题用到前n项大于m的个数。然后用个尺取法。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 200005;
int a[maxn];
int prek[maxn];
int main()
{
    int T;cin>>T;
    while(T--)
    {
        int n,m,k;
        cin>>n>>m>>k;
        memset(prek,0,sizeof(prek));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]>=m) prek[i] += prek[i-1]+1;
            else prek[i] = prek[i-1];
        }
        long long ans = 0;
        int sum = 0;
        int t = 1;
        for(int l=0;l<=n;l++)
        {
            while(t<=n&&sum<k)
            {
                sum = prek[t]-prek[l];
                t++;
            }
            if(sum<k) break;
            ans += (n-t+2);
            sum = prek[t-1]-prek[l+1];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}