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).
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.
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
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+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,k,t,q,a[][maxn],p[maxn];
void init()
{
for(int i=;i<=;i++)
for(int j=;(ll)j+(<<i)-<=n;j++)
a[i][j]=min(a[i-][j],a[i-][j+(<<(i-))]);
}
int get(int l,int r)
{
int x=p[r-l+];
return min(a[x][l],a[x][r-(<<x)+]);
}
int main()
{
int i,j;
for(i=;i<=maxn-;i++)p[i]=+p[i>>];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
rep(i,,n)scanf("%d",&a[][i]);
init();
scanf("%d",&q);
while(q--)
{
int l,r;
scanf("%d%d",&l,&r);
int ans=a[][l],_r=r;
l++;
while()
{
int b=l,r=_r,pos=-;
while(l<=r)
{
int mid=l+r>>;
if(get(b,mid)<=ans)pos=mid,r=mid-;
else l=mid+;
}
if(pos==-)break;
else ans%=a[][pos],l=pos+;
}
printf("%d\n",ans);
}
}
//system("Pause");
return ;
}