Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 2669    Accepted Submission(s): 922 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   Source 2016 ACM/ICPC Asia Regional Dalian Online POINT： 用next数组保存比这个数小的数的位置，在循环找下去，找到最小的就是。 法1 单调栈 法2 两个for循环暴力。 法1: #include <iostream> #include <stdio.h> #include <string.h> #include <stack> using namespace std; int a[100003]; int nxt[100003]; int main() { int n; int T; scanf("%d",&T); while(T--) { stack<int> q; while(!q.empty()) q.pop(); scanf("%d",&n); for(int i=1;i<=n;i++) { nxt[i]=-1; scanf("%d",&a[i]); while(!q.empty()&&a[q.top()]>=a[i]) { nxt[q.top()]=i; q.pop(); } q.push(i); } int m; scanf("%d",&m); while(m--) { int l,r; scanf("%d %d",&l,&r); int now=a[l]; int k=nxt[l]; while(k!=-1&&k<=r) { now=now%a[k]; k=nxt[k]; } printf("%d\n",now); } } } 法2： 就这个片段不一样。 for(int i=1;i<=n;i++) { for(int j=i+1;j<=n;j++) { if(a[i]>=a[j]) { nxt[i]=j; break; } } }