[模拟]subsequence

最新推荐文章于 2019-04-22 19:47:00 发布

原创最新推荐文章于 2019-04-22 19:47:00 发布 · 693 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#La

ACM 专栏收录该内容

62 篇文章

订阅专栏

本文详细介绍了如何通过优化算法解决求解序列中连续子序列和大于等于给定目标值的最短长度的问题。通过引入单调性和二分查找技巧，实现高效求解。实例演示了输入输出格式，并提供了代码实现，帮助读者理解算法思想并进行实际应用。

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integerS (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

Many test cases will be given. 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 there isn't such a subsequence, print 0 on a line by itself.

Sample Input

10 15 
5 1 3 5 10 7 4 9 2 8 
5 11 
1 2 3 4 5

Sample Output

2 
3

容易由单调性想到二分，但是本题的单调性可以继续优化算法。

对于给定i，若j-1，不满足sum[j-1]-sum[i-1] ≥S，则j，一定不满足sum[j]-sum[i-1]≥S，因此局部解区间由(1,n)缩小到(i,n)，即(1,i-1)不需考虑。

#include <cstdio>

int ans = 0x3f3f3f3f;

int min(int a,int b)
{
	if (a < b)
		return a;
	return b;
}

int sum[100010];

int main()
{
	freopen("subsequence.in","r",stdin);
	freopen("subsequence.out","w",stdout);

	int n,s;
	while (scanf("%d%d",&n,&s) == 2)
    {
        ans = 0x3f3f3f3f;
        for (int i=1;i<=n;i++)
        {
            int num;
            scanf("%d",&num);
            sum[i] = sum[i-1]+num;
        }

        int i = 1;
        for (int j=1;j<=n;j++)
        {
            if (sum[j] - sum[i-1] < s)continue;
            while(sum[j]-sum[i]>=s) i++;
            ans = min(ans,j-i+1);
        }
        if (ans == 0x3f3f3f3f)
            printf("0\n");
        else
            printf("%d\n",ans);
    }

	return 0;
}