hdu 5875 2016 ACM/ICPC Asia Regional Dalian Online 1008

本文链接：https://blog.youkuaiyun.com/Ginray/article/details/52496606

本文介绍了一种解决区间取模问题的有效算法。通过使用优先队列预先计算每个元素右侧最近的小于该元素的值，可以高效地处理多个区间内的取模运算。此方法避免了不必要的重复计算，显著提高了计算效率。

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Function

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1427 Accepted Submission(s): 126

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

题意：

给出m个区间（L,R），求出从L处的数左往右取模至R的答案；

思路：

用优先队列求出一个数向右的第一个比他小的值，在求答案的时候就可以直接跳转到比他小的值。

因为在和它取模之后，再和比它大的值取模是没有意义的。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int maxn =100005;
int data[maxn];
int flag[maxn];

priority_queue<int >q;

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        int i,j;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",&data[i]);


        q.push(1);
        for(i=2;i<=n;i++){
            while(!q.empty()&&data[i]<data[q.top()]){
                flag[q.top()]=i;
                q.pop();
            }
            q.push(i);
        }
        while(!q.empty()){
            flag[q.top()]=-1;
            q.pop();
        }

        int m;
        scanf("%d",&m);
        for(i=1;i<=m;i++){
            int l,r;
            scanf("%d%d",&l,&r);
            int fn=l;
            int ans=data[l];
            while(fn<=r){
                int wtf=flag[fn];
                if(wtf==-1)
                    break;
                else if(wtf>r)
                    break;
                else if(wtf<=r){
                    ans=ans%data[wtf];
                    fn=wtf;
                }
                if(ans==1)
                    break;
            }
             printf("%d\n",ans);
            }

        }
        return 0;
    }