POJ_3061_Subsequence

描述

分析

定义区间左右端点l和r.l从1开始循环到n,r向后移动,直到区间和>=s,此时为左端点为l时的最短长度.对于左端点为l+1的情况,使得区间和>s的右端点一定>=r,就让r右移直到满足条件.如果r=n仍无法满足,那对于之后的l都无法满足,即可break.此题用二分O(nlogn),用尺取法O(n).

 1 #include<cstdio>
 2 #include<algorithm>
 3 using std :: min;
 4
 5 const int maxn=100005;
 6 int q,n,s;
 7 int a[maxn];
 8
 9 void solve(int n,int s)
10 {
11     int ans=n+1;
12     int r=1,sum=0;
13     for(int l=1;l<=n;l++)
14     {
15         while(r<=n&&sum<s)
16         {
17             sum+=a[r++];
18         }
19         if(sum<s) break;
20         ans=min(ans,r-l);
21         sum-=a[l];
22     }
23     if(ans>n) ans=0;
24     printf("%d\n",ans);
25 }
26
27 void init()
28 {
29     scanf("%d",&q);
30     while(q--)
31     {
32         scanf("%d%d",&n,&s);
33         for(int i=1;i<=n;i++) scanf("%d",&a[i]);
34         solve(n,s);
35     }
36 }
37
38 int main()
39 {
40     freopen("Subsequence.in","r",stdin);
41     freopen("Subsequence.out","w",stdout);
42     init();
43     fclose(stdin);
44     fclose(stdout);
45     return 0;
46 }