2016大连网络赛 Function

Function

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

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

分析：每次取模第一个比当前答案小于等于的数，RMQ+二分；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t,q,a[30][maxn],p[maxn];
void init()
{
    for(int i=1;i<=29;i++)
        for(int j=1;(ll)j+(1<<i)-1<=n;j++)
            a[i][j]=min(a[i-1][j],a[i-1][j+(1<<(i-1))]);
}
int get(int l,int r)
{
    int x=p[r-l+1];
    return min(a[x][l],a[x][r-(1<<x)+1]);
}
int main()
{
    int i,j;
    for(i=2;i<=maxn-10;i++)p[i]=1+p[i>>1];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        rep(i,1,n)scanf("%d",&a[0][i]);
        init();
        scanf("%d",&q);
        while(q--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            int ans=a[0][l],_r=r;
            l++;
            while(1)
            {
                int b=l,r=_r,pos=-1;
                while(l<=r)
                {
                    int mid=l+r>>1;
                    if(get(b,mid)<=ans)pos=mid,r=mid-1;
                    else l=mid+1;
                }
                if(pos==-1)break;
                else ans%=a[0][pos],l=pos+1;
            }
            printf("%d
",ans);
        }
    }
    //system("Pause");
    return 0;
}