poj 3061 Subsequence（尺取）

最新推荐文章于 2020-05-30 15:12:50 发布

cillyb

最新推荐文章于 2020-05-30 15:12:50 发布

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分类专栏：尺取文章标签：算法尺取

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本文介绍了解决子序列求和问题的两种算法实现方法：前缀和加二分搜索及尺取法。这两种方法分别实现了nlogn和O(n)的时间复杂度。

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Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12626		Accepted: 5331

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

方法一：

可以用前缀和和二分做，复杂度是nlogn

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e5+5;
int a[maxn], sum[maxn];
int main(void)
{
    int t, n, S;
    cin >> t;
    while(t--)
    {
        scanf("%d%d", &n, &S);
        scanf("%d", &a[0]);
        sum[0] = a[0];
        for(int i = 1; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum[i] = sum[i-1]+a[i];
        }
        if(sum[n-1] < S) puts("0");
        else
        {
            int ans = n;
            for(int i = 0; sum[n-1]-sum[i] >= S; i++)
            {
                int t = lower_bound(sum+i, sum+n, S+sum[i]) - sum;
                ans = min(ans, t-i);
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}

方法二：

尺取法o（n）

#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
using namespace std;
const int maxn = 1e5+5;
int a[maxn];
int main(void)
{
    int t, n, S;
    cin >> t;
    while(t--)
    {
        scanf("%d%d", &n, &S);
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        int s = 0, e = 0, sum = 0, ans = n+1;
        while(1)
        {
            while(e < n && sum < S)
                sum += a[e++];
            if(sum < S) break;
            ans = min(ans, e-s);
            sum -= a[s++];
        }
        if(ans > n) puts("0");
        else printf("%d\n", ans);
    }
    return 0;
}