POJ 3061 Subsequence 尺取法

Posted 2020-10-29 caomingpei

题目：

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18528		Accepted: 7921

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

Source

Southeastern Europe 2006

题意：给定一个整数序列，求总和不小于S的最短连续子序列的长度。

思路：两种想法：(1)计算出第i位的前缀和。然后二分查找答案 O(nlogn) (2)尺取法，超过S后，前面开始减小，如果仍然满足条件 更新结果，否则继续向后增加。

代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<string>
 6 #include<vector>
 7 #include<stack>
 8 #include<bitset>
 9 #include<cstdlib>
10 #include<cmath>
11 #include<set>
12 #include<list>
13 #include<deque>
14 #include<map>
15 #include<queue>
16 using namespace std;
17 typedef long long ll;
18 const double PI = acos(-1.0);
19 const double eps = 1e-6;
20 const int INF = 0x3f3f3f3f;
21 
22 const int MAXN = 1e5+10;
23 int a[MAXN];
24 int n = 0;
25 int S = 0;
26 
27 void solve(){
28     int res = n+1;
29     int s = 0;int t= 0;int sum =0;
30     for(;;){
31         while(t<n&&sum<S){
32             sum+=a[t++];
33         }
34         if(sum<S)    break;
35         res = min(res,t-s);
36         sum-=a[s++];
37     }
38     if(res>n){
39         printf("0\n");
40     }else{
41         printf("%d\n",res);
42     }
43 }
44 
45 int main(){
46     int T = 0;
47     scanf("%d",&T);
48     for(int t=0;t<T;t++){
49         scanf("%d %d",&n,&S);
50         for(int i=0;i<n;i++){
51             scanf("%d",&a[i]);
52         }
53         solve();
54     }
55     
56     return 0;
57 }

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