hdu5875 Function (离线处理）

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

 

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

#define LL long long
#define INF 0x3f3f3f3f3f3f3f3f

int T;

int n,m;

int a[120000];
int net[120000];
int pos[120000];
int ans[120000];

int cnt=0;


struct Node
{
    int l,r,id;
} node[120000];


bool cmp(Node a,Node b)
{
    if(a.l!=b.l)
        return a.l<b.l;
    return a.r<b.r;
}

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

            cnt=0;
            a[n+1]=0;
            pos[cnt]=n+1;

            for(int i=n; i>=1; i--)
            {
                while(a[i]<=a[pos[cnt]]) cnt--;

                net[i] = pos[cnt];
                pos[++cnt] = i;
            }

            scanf("%d",&n);

            for(int i=1; i<=n; i++)
            {
                scanf("%d%d",&node[i].l,&node[i].r);
                node[i].id=i;
            }

            sort(node+1,node+n+1,cmp);

            node[0].l=node[0].r=node[0].id=0;

            int pre=0;
            for(int i=1; i<=n; i++)
            {
                if(node[pre].l==node[i].l)
                {
                    int re = ans[node[pre].id];

                    int pp = net[node[pre].r];

                    while(pp <= node[i].r&&re)
                    {
                        re %=a[pp];
                        pp = net[pp];
                    }
                    ans[node[i].id]=re;

                    pre=i;
                }
                else
                {
                    int re = a[node[i].l];
                    int pp = net[node[i].l];

                    while(pp <= node[i].r &&re)
                    {
                        re %= a[pp];
                        pp=net[pp];
                    }
                    ans[node[i].id]=re;

                    pre = i;
                }
            }

            for(int i=1; i<=n; i++)
                printf("%d\n",ans[i]);

        }
    }
    return 0;
}

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