2016 大连网赛---Function(单调栈)

题目链接

http://acm.split.hdu.edu.cn/showproblem.php?pid=5875

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

 

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题意：一个长度为n的数列，q次询问，每次输入l 和 r 表示一个区间A[l]~A[r]  求A[l]%A[l+1]A[l+2]...A[r];

 

思路：a%b=r  那么r<a 所以在区间连续取余中 如果一个数对A[i]取完余，后面的数大于A[i],则不用取余，所以只需要找A[i]后小于A[i]且最近的A[j]进行取余，依次进行下去。其实找距离最近且小于自己的数就是 单调栈，这个算法很神奇，复杂度为O（n）；

 

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
using namespace std;
int A[100005];
int pos[100005];
int s[100005];

int main()
{
    int T,N,M;
    cin>>T;
    while(T--)
    {
        scanf("%d",&N);
        for(int i=1;i<=N;i++)
            scanf("%d",&A[i]);
        int num=1;
        A[N+1]=-1;
        s[1]=N+1;
        for(int i=N;i>=1;i--)///单调栈；
        {
            while(A[i]<A[s[num]]) num--;
            pos[i]=s[num];
            s[++num]=i;
        }
        scanf("%d",&M);
        while(M--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            int tmp=A[l];
            while(l<=r)
            {
                if(tmp==0) break;
                if(pos[l]>r) break;
                tmp=tmp%A[pos[l]];
                l=pos[l];
            }
            printf("%d
",tmp);
        }
    }
    return 0;
}