题目:点击打开链接
题意:给出一个序列,要求的是一个区间,这个区间的最小值乘以这个区间数字的和 是最大值。求这个最大值与这个区间。
分析:单调队列经典题,在以每个数为最小值的情况中,取个最大值,维护每个数可以左右延伸到的位置,预处理前缀和,复杂度O(n)。实现细节见代码。
代码:
#pragma comment(linker, "/STACK:102400000,102400000")///手动扩栈
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
typedef pair<int,int> PII;
const ll mod = 1e9+7;
const int N = 1e6+10;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int n,a[N],q[N],l,r,lf[N],top;
ll s[N],ans,tp;
void sv() {
top=0,ans=-1;
for(int i=1;i<=n;i++) {
if(top==0 || a[i]>a[q[top-1]]) {
lf[i]=i;
q[top++]=i;
continue;
}
if(a[i]==a[q[top-1]]) continue;
while(top && a[i]<a[q[top-1]]) {
top--;
tp = 1LL*a[q[top]]*(s[i-1]-s[lf[q[top]]-1]);
if(tp>ans) {
l = lf[q[top]];
r = i - 1;
ans = tp;
}
}
lf[i] = lf[q[top]];
q[top++]=i;
}
cout<<ans<<endl<<l<<" "<<r<<endl;
}
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
while(cin>>n) {
for(int i=1;i<=n;i++) cin>>a[i],s[i] = s[i-1] + a[i];
a[++n]=-1;
sv();
mst(s,0);
}
return 0;
}