2016大连网络赛 Function

Posted 2020-08-08 mxzf0213

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：暴力转移即可；（此方法复杂度高，待改进，不过能A。。。）

　　　参考http://blog.csdn.net/angon823/article/details/52496697

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[maxn],r[maxn];
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        rep(i,1,n)scanf("%d",&a[i]);
        memset(r,inf,sizeof r);
        r[n]=inf;
        for(i=n-1;i>=1;i--)
        {
            k=i+1;
            while(k<=n)
            {
                if(a[i]>a[k])
                {
                    r[i]=k;
                    break;
                }
                k=r[k];
            }
        }
        scanf("%d",&q);
        while(q--)
        {
            int x,y,ans;
            scanf("%d%d",&x,&y);
            ans=a[x];
            while(r[x]<=y)
            {
                ans%=a[r[x]];
                x=r[x];
            }
            printf("%d\n",ans);
        }
    }
    //system("Pause");
    return 0;
}