uva1619

分析：这个题的关键是要找到，当某个值是最小值时它最大的影响区间时什么。可以通过单调队列（单调栈）在nlogn的时间内实现

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <queue>
 using namespace std;
 const int maxn=+;
 int n;
 int a[maxn],L[maxn],R[maxn];
 long long sum[maxn];
 int main(){
     int t=;
     //freopen("out.txt","w",stdout);
     while(scanf("%d",&n)!=EOF){
         if(t)printf("\n");
         t++;
         sum[]=;
         for(int i=;i<=n;i++){
             scanf("%d",&a[i]);
             sum[i]=sum[i-]+a[i];
         }
       deque<int>q;
       for(int i=;i<=n;i++){
         while(!q.empty()&&a[i]<=a[q.back()]){
             R[q.back()]=i-;
             q.pop_back();
         }
         if(q.empty())L[i]=;
         else L[i]=q.back()+;
         q.push_back(i);
       }
       while(!q.empty()){
         R[q.back()]=n;
         q.pop_back();
       }
       long long ans=;
       int ansL=,ansR=;
       for(int i=;i<=n;i++){
         if(ans<(long long)(sum[R[i]]-sum[L[i]-])*a[i]){
             ans=(long long)(sum[R[i]]-sum[L[i]-])*a[i];
             ansL=L[i],ansR=R[i];
         }
       }
       printf("%lld\n",ans);
       printf("%d %d\n",ansL,ansR);
     }
 return ;
 }