HDU 5875 Function 大连网络赛 线段树

Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1498    Accepted Submission(s): 553

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


函数的意思解读出来就是在l到r的区间里，a[l],对区间里的数，逐个取余
那么比a[l]大的数，取余不变，主要看比a[l]小的数，所以在区间里找第一个比a[l]小的数，
然后继续在剩下的区间里面找比取完余的a[l]小的数

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <string>
#include <stdlib.h>
#include <vector>

using namespace std;
const int maxn=1e5;
int cmin[maxn*4+5];
int a[maxn+5];
int n,m;
int l,r;
int x;
void PushUp(int node)
{
    cmin[node]=min(cmin[node<<1],cmin[node<<1|1]);
}
void build(int node,int begin,int end)
{
    if(begin==end)
    {
         scanf("%d",&x);
         cmin[node]=x;
         a[begin]=x;
         return;
    }
    int m=(begin+end)>>1;
    build(node<<1,begin,m);
    build(node<<1|1,m+1,end);
    PushUp(node);
}
bool tag;
int minn;
int pos;
void query(int node,int begin,int end,int left,int right,int value)
{
    if(value<cmin[node])
        return;

    if(begin==end)
    {
        minn=cmin[node];
        pos=begin;
        tag=true;
        return;
    }

    int m=(begin+end)>>1;
    if(left<=m)
        query(node<<1,begin,m,left,right,value);
    if(tag) return;
    if(right>m)
        query(node<<1|1,m+1,end,left,right,value);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        build(1,1,n);
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&l,&r);
            tag=false;
            int x=a[l];
            if(l==r)
            {
                printf("%d\n",x);
                continue;
            }
            query(1,1,n,l+1,r,x);
            if(!tag)
                printf("%d\n",x);
            else
            {
                while(1)
                {
                    tag=false;
                    x%=minn;
                    if(pos+1>r)
                    {
                        printf("%d\n",x);
                        break;
                    }
                    query(1,1,n,pos+1,r,x);
                    if(!tag)
                    {
                        printf("%d\n",x);
                        break;
                    }
                }
            }

        }
    }
    return 0;
}