HDOJ5875(线段树)

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 l, r, mn;
}a[MAXN*3];
int n, m, val[MAXN];
void build(int rt, int l, int r)
{
    a[rt].l = l;
    a[rt].r = r;
    if(l == r)
    {
        scanf("%d", &val[l]);
        a[rt].mn = val[l];
        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 &x)
{
    if(x == 0)  return ;
    if(a[rt].mn > x)    return ;
    if(a[rt].l == a[rt].r)
    {
        x %= a[rt].mn;
        return ;
    }
    int mid = (a[rt].l + a[rt].r) >> 1;
    if(r <= mid)
    {
        query(rt << 1, l, r, x);
    }
    else if(mid < l)
    {
        query((rt << 1) | 1, l, r, x);
    }
    else
    {
        query(rt << 1, l, mid, x);
        query((rt << 1) | 1, mid + 1, r, x);
    }
}
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 x = val[l];
                query(1, l + 1, r, x);
                printf("%d\n", x);
            }
        }
    }
    return 0;
}