尺取法（滑窗法）的应用（POJ 3061,POJ 3320)

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

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

23

当题目中涉及连续一段的大于某值而且求距离最短可以考虑尺取法或者说是

滑窗法：具体就是两个端点，左边端点按次加，右边滑动找满足条件的点。

更新最小值。复杂度o(n).

但是这也不是唯一方法。同样我们可以统计前缀和，然后求右端点二分。

这样一来复杂度应该是o(nlogn)

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int maxn=1e5+10;int t,n;LL m,a[maxn];void solve(){    int st=0,ed=0;    LL sum=0;    int res=n+1;    while(1)    {        while(sum<m&&ed<n)            sum+=a[ed++];        if(sum<m)  break;        res=min(res,ed-st);        sum-=a[st];st++;    }    if(res==n+1)  res=0;    printf("%d\n",res);}int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d%lld",&n,&m);        REP(i,n)  scanf("%lld",&a[i]);        solve();    }    return 0;}

Jessica's Reading Problem

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7512 Accepted: 2387

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

51 8 8 8 1

Sample Output

2

Source

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int maxn=1e6+10;set<int>s;map<int,int>p;int a[maxn];int n;void solve(){    int st=0,ed=0,num=0;    int res=n;int dd=s.size();    while(1)    {        while(ed<n&&num<dd)        {            if(!p[a[ed]]) num++;            p[a[ed]]++;ed++;        }        if(num<dd)  break;        res=min(res,ed-st);        p[a[st]]--;if(!p[a[st]]) num--;        st++;    }    printf("%d\n",res);}int main(){    while(~scanf("%d",&n))    {        s.clear();p.clear();        REP(i,n)        {            scanf("%d",&a[i]);            s.insert(a[i]);        }        solve();    }    return 0;}