The shorter, the simpler. With this problem, you should be convinced of this truth.
You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1≤l≤r≤N)F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.F(l,r)={All=r;F(l,r−1) modArl<r.
You job is to calculate F(l,r)F(l,r), for each query (l,r)(l,r).
You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1≤l≤r≤N)F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.F(l,r)={All=r;F(l,r−1) modArl<r.
You job is to calculate F(l,r)F(l,r), for each query (l,r)(l,r).
InputThere are multiple test cases.
The first line of input contains a integer TT, indicating number of test cases, and TT test cases follow.
For each test case, the first line contains an integer N(1≤N≤100000)N(1≤N≤100000).
The second line contains NN space-separated positive integers: A1,…,AN (0≤Ai≤109)A1,…,AN (0≤Ai≤109).
The third line contains an integer MM denoting the number of queries.
The following MM lines each contain two integers l,r (1≤l≤r≤N)l,r (1≤l≤r≤N), representing a query.OutputFor each query(l,r)(l,r), output F(l,r)F(l,r) on one line.
Sample Input
1 3 2 3 3 1 1 3
Sample Output
2
题意：
已知a[]数组，现在给出m组l,r。求a[l]%a[l+1]%a[l+2]%...a[r]的结果。
思路：
a[l]<a[l+1]，那么这个膜运算是可以忽视的。单调队列，逆序预处理出每一个a[]的右边第一个小于a[]的数的位置R[]，然后就是用a[l]%a[R[l]]%a[R[R[l]]]...。
#include<cstdio> #include<cstdlib> #include<iostream> #include<cstring> #include<algorithm> using namespace std; const int maxn=1000010; int a[maxn],R[maxn]; int main() { int n,m,T,i,j,l,r,t,ans; scanf("%d",&T); while(T--){ scanf("%d",&n); for(i=1;i<=n;i++) scanf("%d",&a[i]); for(i=1;i<=n+1;i++) R[i]=n+1; a[n+1]=0; for(i=n-1;i>=1;i--){ t=i; while(t<n){ // 得到R数组 if(a[t+1]<a[i]){ R[i]=t+1; break;} if(a[R[t+1]]>=a[i]) t=R[t+1]; else { for(j=t+1;j<=R[t+1];j++) {//这里可以二分优化 if(a[j]<a[i]) { R[i]=j;break; } } break; } } } scanf("%d",&m); while(m--){ scanf("%d%d",&l,&r); ans=a[l];t=l; while(R[t]<=r){ //过滤掉无用的膜运算 t=R[t]; ans%=a[t]; } printf("%d\n",ans); } } return 0; }
