poj 3061 Subsequence

最新推荐文章于 2021-08-10 16:08:42 发布

abns

最新推荐文章于 2021-08-10 16:08:42 发布

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分类专栏： poj 二分尺取法

本文链接：https://blog.youkuaiyun.com/qq_41453511/article/details/81706761

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本文介绍了一种算法，用于寻找给定整数序列中最短的连续子序列，该子序列的元素之和大于或等于指定的目标值。通过两种实现方式——滑动窗口和二分搜索，来解决这一问题。

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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

二分尺取都可以

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<math.h>
#define inf 0x3f3f3f3f
#define maxn 3000000
using namespace std;
int a[maxn];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int l=0,r=0,sum=0,ans=inf;
        while(1)
        {
            while(sum<m&&r<n)
            {
                    sum+=a[r++];

            }
            if(sum<m)break;
            ans=min(ans,r-l);
            //cout<<r-l<<endl;
            sum-=a[l++];
        }
        if(ans==inf) ans=0;
        printf("%d\n",ans);
    }
	return 0;
}

二分

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<math.h>
#define inf 0x3f3f3f3f
#define maxn 3000000
using namespace std;
int a[maxn];
int b[maxn];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=b[i-1]+a[i];
        }
        int ans=inf;
        for(int i=1;b[n]-b[i]>=m;i++)
        {
            int temp=lower_bound(b+1+i,b+n+1,b[i]+m)-(b+i);
            ans=min(temp,ans);
        }
         if(ans==inf) ans=0;
         printf("%d\n",ans);
    }
	return 0;
}