HDOJ5875(线段树)

Posted 2020-08-09 vCoders

Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1701    Accepted Submission(s): 615

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

思路：用val[l],每次%上在[l+1,r]中下一个最小的值。

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN=100005;
struct Node{
    int mn,l,r;
}a[MAXN*3];
int n,m;
int val[MAXN];
void build(int rt,int l,int r)
{
    a[rt].l=l;
    a[rt].r=r;
    if(l==r)
    {
        scanf("%d",&a[rt].mn);
        val[l]=a[rt].mn;
        return ;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build((rt<<1)|1,mid+1,r);
    a[rt].mn=min(a[rt<<1].mn,a[(rt<<1)|1].mn);
}
void query(int rt,int l,int r,int &val)
{
    if(val==0)    return ;
    if(a[rt].mn>val)    return ;
    if(a[rt].l==a[rt].r)
    {
        val%=a[rt].mn;
        return ;
    }
    int mid=(a[rt].l+a[rt].r)>>1;
    if(r<=mid)
    {
        query(rt<<1,l,r,val);
    }
    else if(mid<l)
    {
        query((rt<<1)|1,l,r,val);
    }
    else
    {
        query(rt<<1,l,mid,val);
        query((rt<<1)|1,mid+1,r,val);
    }
} 
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);    
        build(1,1,n);
        scanf("%d",&m);
        while(m--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            if(l==r)
            {
                printf("%d\n",val[l]);
            }
            else
            {
                int res=val[l];
                query(1,l+1,r,res);
                printf("%d\n",res);
            }    
        }
    }
    return 0;
}