POJ 3061 Subsequence

Subsequence
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 7307   Accepted: 2710

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

 

首先用一个数组sum[i]维护一个前缀和，sum[0]=0，sum[i]=sum[1]+sum[2]+……+sum[i]；

然后可知区间(s,t]的和就是sum[t]-sum[s]；

之后可以枚举所有满足sum[t]>s的t，对每个t用二分查找的方法在前面找到满足条件且与它距离最近的s，求出最小值即可

 

 

 1 #include<iostream>
 2 #include<cstdio>
 3 
 4 using namespace std;
 5 
 6 int n,s;
 7 int sum[100010];
 8 
 9 int main()
10 {
11     int kase;
12 
13     scanf("%d",&kase);
14 
15     while(kase--)
16     {
17         int temp,t=-1;
18         sum[0]=0;
19         scanf("%d %d",&n,&s);
20         for(int i=1;i<=n;i++)
21         {
22             scanf("%d",&temp);
23             sum[i]=sum[i-1]+temp;
24             if(t==-1&&sum[i]>=s)
25                 t=i;
26         }
27         if(t==-1)
28         {
29             printf("0\n");
30             continue;
31         }
32         int ans=n;
33         for(;t<=n;t++)
34         {
35             int l=0,r=t-1;
36             while(r-l>1)
37             {
38                 int mid=(r+l)/2;
39                 if(sum[t]-sum[mid]>=s)
40                     l=mid;
41                 else
42                     r=mid-1;
43             }
44             if(sum[t]-sum[r]>=s)
45                 l=r;
46             if(t-l<ans)
47                 ans=t-l;
48         }
49         printf("%d\n",ans);
50     }
51 
52     return 0;
53 }
[C++]

 

 在《挑战程序设计竞赛》书上看到了用尺取法优化到O(n)的做法，例如Sample Input中第一组数据：

5  1  3  5  10  7  4  9  2  8

s               t                          ans=5

5  1  3  5  10  7  4  9  2  8

    s           t                          ans=4

5  1  3  5  10  7  4  9  2  8

        s       t                          ans=3

5  1  3  5  10  7  4  9  2  8

           s    t                          ans=2

5  1  3  5  10  7  4  9  2  8

                s    t                     ans=2

5  1  3  5  10  7  4  9  2  8

                     s      t              ans=2

5  1  3  5  10  7  4  9  2  8

                         s      t          ans=2

5  1  3  5  10  7  4  9  2  8

                             s      t      ans=2

可知最终答案是2

 

 

 1 #include<iostream>
 2 #include<cstdio>
 3 
 4 using namespace std;
 5 
 6 int n,s;
 7 int sum[100010];
 8 
 9 int main()
10 {
11     int kase;
12 
13     scanf("%d",&kase);
14 
15     while(kase--)
16     {
17         int temp,l=0,r=-1;
18         sum[0]=0;
19         scanf("%d %d",&n,&s);
20         for(int i=1;i<=n;i++)
21         {
22             scanf("%d",&temp);
23             sum[i]=sum[i-1]+temp;
24             if(r==-1&&sum[i]>=s)
25                 r=i;
26         }
27         if(r==-1)
28         {
29             printf("0\n");
30             continue;
31         }
32         int ans=n;
33         while(true)
34         {
35             if(sum[r]-sum[l]>=s)
36             {
37                 if(r-l<ans)
38                     ans=r-l;
39                 l++;
40             }
41             else
42             {
43                 if(r==n)
44                     break;
45                 r++;
46             }
47         }
48         printf("%d\n",ans);
49     }
50 
51     return 0;
52 }
[C++]

 

posted @ 2013-09-04 10:55  ~~Snail~~  阅读(272)  评论(0)  编辑  收藏  举报