HDU 5875 Function (单调栈+暴力)

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本文介绍了一种数组查询问题，给定一个正整数数组和多个查询，每个查询指定两个索引(l,r)，任务是计算从l到r的特定函数F(l,r)。提供了两种解决方法：一种使用单调栈来预处理数组，另一种采用简单的暴力搜索。

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Function

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2669 Accepted Submission(s): 922

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

Sample Output

Source

2016 ACM/ICPC Asia Regional Dalian Online

POINT：

用next数组保存比这个数小的数的位置，在循环找下去，找到最小的就是。

法1 单调栈法2 两个for循环暴力。

法1:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;
int a[100003];
int nxt[100003];
int main()
{
    int n;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        stack<int> q;
        while(!q.empty()) q.pop();
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            nxt[i]=-1;
            scanf("%d",&a[i]);
            while(!q.empty()&&a[q.top()]>=a[i])
            {
                nxt[q.top()]=i;
                q.pop();
            }
            q.push(i);
        }
        
        int m;
        scanf("%d",&m);
        while(m--)
        {
            int l,r;
            scanf("%d %d",&l,&r);
            int now=a[l];
            int k=nxt[l];
            while(k!=-1&&k<=r)
            {
                now=now%a[k];
                k=nxt[k];
            }
            printf("%d\n",now);
            
        }

    }
    
    
    
}

法2：就这个片段不一样。

        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(a[i]>=a[j])
                {
                    nxt[i]=j;
                    break;
                }
            }
        }