poj3061 Subsequence(尺取法)

https://vjudge.net/problem/POJ-3061

尺取发,s和t不断推进的算法。因为每一轮s都推进1所以复杂度为O(n)

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<stack>
 8 #define lson l, m, rt<<1
 9 #define rson m+1, r, rt<<1|1
10 #define INF 0x3f3f3f3f
11 typedef unsigned long long ll;
12 using namespace std;
13 int a[100010];
14 int main()
15 {
16     int kase, n, m, S;
17     cin >> kase;
18     while(kase--){
19         cin >> n >> S;
20         for(int i = 0; i < n; i++){
21             cin >> a[i];
22         }
23         int s=0, t=0, sum=0, ans = INF;//尺取法,开始s和结尾t不断推进 
24         while(1){
25             while(t < n&&sum < S){
26                 sum += a[t++];
27             }
28             if(sum<S) break;//也即t>=n退出 
29             ans = min(ans, t-s);
30             sum -= a[s++];//s推进,sum修改 
31         }
32         if(ans == INF) //没有符合要求的 
33             cout << "0" << endl;
34         else cout << ans << endl;
35     } 
36     return 0;
37 }
原文地址:https://www.cnblogs.com/Surprisezang/p/9017818.html