HDU5875-Function

最新推荐文章于 2021-03-31 20:39:55 发布

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本文介绍了一种区间取余查询问题的解决方法，通过维护单调递增序列和使用单调栈来提高查询效率。针对给定的整数数组，实现快速查询指定区间内元素依次取余的结果。

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Function

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 851 Accepted Submission(s): 321

Problem Description

The shorter, the simpler. With this problem, you should be convinced of this truth.

You are given an array

A of

N postive integers, and

M queries in the form

(l,r). A function

F(l,r) (1≤l≤r≤N) is defined as:

F(l,r)={AlF(l,r−1) modArl=r;l<r.
You job is to calculate

F(l,r), for each query

(l,r).

Input

There are multiple test cases.

The first line of input contains a integer

T, indicating number of test cases, and

T test cases follow.

For each test case, the first line contains an integer

N(1≤N≤100000).
The second line contains

N space-separated positive integers:

A1,…,AN (0≤Ai≤109).
The third line contains an integer

M denoting the number of queries.
The following

M lines each contain two integers

l,r (1≤l≤r≤N), representing a query.

Output

For each query

(l,r), output

F(l,r) on one line.

Sample Input

  1 3 2 3 3 1 1 3
 

Sample Output

2

Source

2016 ACM/ICPC Asia Regional Dalian Online

题意：给你n个数，给定区间[l,r],求a[l] % a[l + 1] % ... % a[r]

解题思路：对于%，只有比本身小到数才会影响最后结果，所以我们维护一个单调递增序列，和next数组，表示比当前数小到下一个数的位置，采用单调栈或者暴力都可以

暴力：

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int main()
{
    int t,n,m,l,r,a[100009],next[100009];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        memset(next,-1,sizeof next);
        for(int i=1; i<=n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                if(a[j]<=a[i])
                {
                    next[i]=j;
                    break;
                }
            }
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d %d",&l,&r);
            int x=a[l];
            for(int i=next[l]; i<=r; i=next[i])
            {
                if(i==-1) break;
                x%=a[i];
            }
            printf("%d\n",x);
        }
    }
    return 0;
}

单调栈：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, a[1000090], nt[1000009];

int main()
{
	int t;
	while (~scanf("%d", &t))
	{
		while (t--)
		{
			scanf("%d", &n);
			memset(nt, -1, sizeof nt);
			stack<int>s;
			for (int i = 1; i <= n; i++)
			{
				scanf("%d", &a[i]);
				while (!s.empty() && a[s.top()] > a[i]) nt[s.top()] = i, s.pop();
				s.push(i);
			}
			int m;
			scanf("%d", &m);
			while (m--)
			{
				int l, r;
				scanf("%d%d", &l, &r);
				int ans = a[l], k = l;
				while (nt[k] <= r&&nt[k] != -1)
				{
					ans %= a[nt[k]];
					k = nt[k];
				}
				printf("%d\n", ans);
			}
		}
	}
	return 0;
}