poj2796(单调栈)

传送门

实测如果有多个值相等，输出最后一个最大值的LR

ac代码：

#include<bits/stdc++.h>
#define per(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;
//#define int long long
const ll inf =2333333333333333LL;
const double eps=1e-8;
int read(){
    char ch=getchar();
    int res=0,f=0;
    while(ch<'0' || ch>'9'){f=(ch=='-'?-1:1);ch=getchar();}
    while(ch>='0'&&ch<='9'){res=res*10+(ch-'0');ch=getchar();}
    return res*f;
}
// ------------------------head
#define mod 1000000007
const int N=100005;
int n,L[N],R[N];
ll s[N],a[N];
stack<ll>st;

signed main()
{
    while(scanf("%d",&n)!=EOF){
        s[0]=0;
        while(!st.empty())st.pop();
        per(i,1,n){scanf("%lld",&a[i]);s[i]=s[i-1]+a[i];}
        for(int i=1;i<=n;i++){
            while(!st.empty()&&a[st.top()]>=a[i])st.pop();
            if(st.empty())L[i]=0;
            else L[i]=st.top();
            st.push(i);
        }
        while(!st.empty())st.pop();
        for(int i=n;i>=1;i--){
            while(!st.empty()&&a[st.top()]>=a[i])st.pop();
            if(st.empty())R[i]=n;
            else R[i]=st.top()-1;
            st.push(i);
        }
        ll res=0;
        int k;
        per(i,1,n){
            ll tmp=(ll)a[i]*(s[R[i]]-s[L[i]]);
            if(tmp>=res){
                res=tmp;
                k=i;
            }
        }
        printf("%lld\n%d %d\n",res,L[k]+1,R[k]);
    }
    return 0;
}