Subsequence（二分）

最新推荐文章于 2025-01-10 21:54:54 发布

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本文介绍了一个寻找最小子序列的问题，给定一系列正整数及目标值S，通过编程找到连续子序列的最小长度，该子序列的和大于等于S。使用C++实现，通过二分查找优化解题过程。

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

一开始c++交不过，g++过，原来是int 输入的时候是long long，wa了，g++交没这个问题

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
ll a[maxn];
ll s[maxn];
ll n,ss;
bool check(int x)
{
    if(x>=n)
    {
    	return true;
	}
	for(int t=x;t<=n;t++)
	{
//		cout<<s[t]<<" "<<s[t-x+1]<<endl;
		if(s[t]-s[t-x+1]>=ss)
		{
			
			return true;
		}
	}
	return false;
}
int  main()
{
	int  T;
	cin>>T;
	while(T--)
	{
		scanf("%lld%lld",&n,&ss);
		for(int t=1;t<=n;t++)
		{
			scanf("%lld",&a[t]);
		}
		memset(s,0,sizeof(s));
		for(int t=1;t<=n;t++)
		{
			s[t]=s[t-1]+a[t];
		}
		if(s[n]<ss)
		{
			cout<<"0"<<endl;
	    }
	    else
	    {
		int l=0,r=100000;
		int mid;
		while(l<=r)
		{
			mid=(l+r)/2;
			if(check(mid))
			{
		      r=mid-1;
			}
			else
			{
			  l=mid+1;
			}
		}
		mid=(l+r)/2;
		cout<<mid<<endl;
	   }
	}
	
	return 0;
}