看了大佬的博客，又做了几道题，对尺取法有了一点感觉。

首先，什么情况下能使用尺取法？尺取法通常适用于选取区间有一定规律，或者说所选取的区间有一定的变化趋势的情况，通俗地说，在对所选取区间进行判断之后，我们可以明确如何进一步有方向地推进区间端点以求解满足条件的区间，如果已经判断了目前所选取的区间，但却无法确定所要求解的区间如何进一步

得到根据其端点得到，那么尺取法便是不可行的。首先，明确题目所需要求解的量之后，区间左右端点一般从最整个数组的起点开始，之后判断区间是否符合条件在根据实际情况变化区间的端点求解答案。下面是几道尺取法的题。

Subsequence

Description

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

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int n,s,t,arr[100001];
int main()
{
    scanf("%d",&t);
    while(t)
    {
        scanf("%d%d",&n,&s);
        for(int i = 1;i <= n;++i)
            scanf("%d",&arr[i]);
        int j = 1,sum = 0,ans = INT_MAX;
        for(int i = 1;i <= n;++i)
        {
            sum += arr[i];
            while(sum >= s)
            {
                ans = min(ans,i - j + 1);
                sum -= arr[j];
                ++j;
            }
        }
        if(ans == INT_MAX)
            printf("%d\n",0);
        else
            printf("%d\n",ans);
        --t;
    }
    return 0;
}

Jessica’s Reading Problem

Description

Jessica’s a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that text book, like other authors, is extremely fussy about the ideas, thus some ideas are covered more than once. Jessica think if she managed to read each idea at least once, she can pass the exam. She decides to read only one contiguous part of the book which contains all ideas covered by the entire book. And of course, the sub-book should be as thin as possible.

A very hard-working boy had manually indexed for her each page of Jessica’s text-book with what idea each page is about and thus made a big progress for his courtship. Here you come in to save your skin: given the index, help Jessica decide which contiguous part she should read. For convenience, each idea has been coded with an ID, which is a non-negative integer.

Input

The first line of input is an integer P (1 ≤ P ≤ 1000000), which is the number of pages of Jessica’s text-book. The second line contains P non-negative integers describing what idea each page is about. The first integer is what the first page is about, the second integer is what the second page is about, and so on. You may assume all integers that appear can fit well in the signed 32-bit integer type.

Output

Output one line: the number of pages of the shortest contiguous part of the book which contains all ideals covered in the book.

Sample Input

5
1 8 8 8 1

Sample Output

2

这一道题也差不多，只不过32位整数太大了，就用map，思路和上一题差不多

#include<map>
#include<set>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include<iostream>
using namespace std;
int p,arr[1000001];
set<int> s;
map<int,int> m;
int main()
{
    scanf("%d",&p);
    for(int i = 1;i <= p;++i)
    {
        scanf("%d",&arr[i]);
        s.insert(arr[i]);
    }
    int sum = s.size(),num = 0,j = 1,ans = INT_MAX;
    for(int i = 1;i <= p;++i)
    {
        int temp = m[arr[i]];
        if(!temp)
            ++num;
        m[arr[i]] = ++temp;
        while(num == sum)
        {
            ans = min(ans,i - j + 1);
            int temp2 = m[arr[j]];
            m[arr[j]] = --temp2;
            if(!temp2)
                --num;
            ++j;
        }
    }
    printf("%d\n",ans);
    return 0;
}