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最新推荐文章于 2025-08-05 19:50:13 发布

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最新推荐文章于 2025-08-05 19:50:13 发布

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CC 4.0 BY-SA版权

文章标签：算法

本文链接：https://blog.youkuaiyun.com/qq_63010655/article/details/123668737

这篇博客介绍了如何利用前缀和结合二分查找方法来求解区间相加和的问题。当给定一个整数数组和一个目标值，目标是找到数组中连续子数组的最小长度，使得子数组之和大于或等于目标值。文章通过两个不同的实现方式，一种是二分查找，另一种是双指针，详细解释了算法思路并提供了C++代码示例。

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求区间相加和，可以用前缀和，在加二分的方法，可以找到最小的满足条件的值

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 10;
typedef long long ll;
ll n, h;
ll a[N];
ll cha;
ll minn = 1e18;//代表两者之间的最小距离 

int check(ll l, ll r)
{
//	cout << "l:" << l << "r:" << r << endl;
	if (a[r] - a[l - 1] >= cha)
	{
		if (a[r] - a[l - 1] < minn)
		{
			minn = a[r] - a[l - 1];
		}
		return 1;
	}
	return 0;
}

void erfendistance(ll temp)
{
	ll l = temp + 1, r = n, mid;
//	cout << "l:" << l << "r:" << r << endl;
	while (l <= r)
	{
		mid = (l + r) / 2;
		if (check(temp, mid))
			r = mid - 1;
		else
			l = mid + 1;	
	}
} 

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cin >> n >> h;
	
	for (ll i = 1; i <= n; ++i)
	{
		cin >> a[i];
		a[i] = a[i] + a[i - 1];
	}
	
	if (a[n] < h)
		cout << h - a[n] << endl;
	else if (n == 1)
	{
		cout << h << endl;
	}		
	else 
	{
		cha = a[n] - h + 1;
		for (ll i = 1; i <= n; ++i)
		{
			erfendistance(i);
		}
		cout << h - (a[n] - minn) << endl;
	}	
	return 0;
}

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 10;
typedef long long ll;
ll n, h;
ll a[N];
ll minn = 1e18;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cin >> n >> h;

	for (ll i = 1; i <= n; ++i)
	{
		cin >> a[i];
		a[i] = a[i] + a[i - 1];
	} 
	
	if (a[n] < h)
	{
		cout << h - a[n] << endl;
	}
	else if (n == 1)
	{
		cout << h << endl;
	}
	else
	{
		ll cnt = 1, flag = 0;
		ll cha = a[n] - h + 1;
		for (ll i = 1; i <= n; ++i)
		{
			if (cnt == n && a[cnt] - a[i] < cha)
				break;
			for (ll j = cnt; j <= n;)
			{
				if (a[j] - a[i - 1] >= cha)
				{
					if (a[j] - a[i - 1] < minn)
					{
						minn = a[j] - a[i - 1];
					}
					break;
				}
				else
				{
					cnt++;
					j++;
				}
			}
		}
		
		cout << h - (a[n] - minn) << endl;
	}
	
	return 0;
}

一种是二分，一种是双指针