尺取

最新推荐文章于 2023-02-06 21:56:34 发布

原创最新推荐文章于 2023-02-06 21:56:34 发布 · 393 阅读

0 ·

CC 4.0 BY-SA版权

尺取专栏收录该内容

2 篇文章

订阅专栏

本文解析了两道经典的尺取法编程题目：一是寻找最短子序列使其元素之和大于等于给定值；二是计算包含特定种类字符的子串数量。通过示例代码详细介绍了尺取法的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1.sequence

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

Input

The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

Output

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

Sample Input

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

Sample Output

2
3

题意:

给定一个序列a[]和一个maxs,求不小于maxs的子序列的长度最小值

解析:

最简单尺取

求从左到右得一个>=maxs的子序列,然后减掉末尾的呢个,再看前面需要加几个(包括0)可以和>=maxs,直到前面加不了为止,结束

ac:

#include<cstdio>
#include<algorithm>
#define ll long long
#define MAXN 100005
using namespace std;
int a[MAXN];

int main()
{
    int t,maxs,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&maxs);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        ll sum=0;
        int i=0,j=0;
        int ans=99999999;
        while(1)
        {
            while(sum<maxs&&j<n)
                sum+=a[j],j++;
            if(sum<maxs)
                break;
            ans=min(ans,j-i);
            sum-=a[i],i++;
        }
        if(ans==99999999)
            printf("0\n");
        else printf("%d\n",ans);
    }
    return 0;
}

S - String

HDU - 5672

题意:

给一个字符串,一个n,求存在多少个子串至少包含了 kk 种子母

解析:

尺取法,要保存不同字母的数目

ac:

#include<bits/stdc++.h>
#define ll long long
#define MAXN 1000005
using namespace std;
char str[MAXN];
int vis[27]={0};
set<int> st;

int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        st.clear();
        scanf("%s",&str);
        int len=strlen(str);
        scanf("%d",&n);
        int i=0,j=0;
        ll sum=0;
        while(1)
        {
            while(st.size()<n&&j<len)
            {
                st.insert(str[j]-'a');
                vis[str[j]-'a']++;
                j++;
            }
            if(st.size()<n)
                break;
            if(st.size()==n)
                sum+=len-j+1;
            vis[str[i]-'a']--;
            if(vis[str[i]-'a']==0)
                st.erase(str[i]-'a');
            i++;
        }
        printf("%lld\n",sum);
    }
    return 0;
}