标签：tput integer question open acm ber core and The
SOJ 3085: windy‘s cake V http://acm.scu.edu.cn/soj/problem.action?id=3085
Problem: Given a list of $n$ positive integers $num[1], ..., num[n]$, denote the score of one sublist from $i$ to $j$ as $(\min_{i\le k\le j}num[k])*\left(\Sigma_{k=i}^{j}num[k]\right)$. The question is that which sublist has the largest score? Output the largest score.
Analysis: we can enumerate each $num[i], 1\le i\le n$ to be the minimum number in one sublist. Therefore, for each $num[i], 1\le i\le n$, we need to find the corresponding sublist, that is, to find the first number $num[i‘]$ that is smaller than $num[i]$ from $i-1$ to $1$ and the first number $num[i‘‘]$ that is smaller than $num[i]$ from $i+1$ to $n$ (similar to SOJ 2511, please see here). Then, for $num[i]$ the score $score[i]=num[i]*\left(\Sigma_{k=i‘}^{i‘‘}num[k]\right)$, and the solution is $\max_{1\le k\le n}score[i]$.
Technique: Maintaining an increasing stack.
Code:
#include<iostream>
#include<stack>
using namespace std;
struct node
{
int no;
int num;
int prevNo;
node(){prevNo=0;}
};
node arr[100005];
stack<node>s;
long long sum[100005];
int main()
{
int n;
int temp1;
int i;
long long temp2;
long long ans;
while(scanf("%d",&n)==1)
{
sum[0]=0;
for(i=1;i<=n;i++)
{
scanf("%d",&temp1);
sum[i]=sum[i-1]+temp1;
arr[i].no=i;
arr[i].num=temp1;
}
arr[0].no=0;
arr[0].num=-1;
arr[n+1].no=n+1;
arr[n+1].num=-1;
s.push(arr[0]);
ans=0;
for(i=1;i<=n+1;i++)
{
while(arr[i].num<s.top().num)
{
temp2=s.top().num*(sum[i-1]-sum[s.top().prevNo]);
ans=temp2>ans ? temp2 : ans;
s.pop();
}
arr[i].prevNo=s.top().no;
s.push(arr[i]);
}
while(!s.empty())
s.pop();
printf("%lld\n",ans);
}
return 0;
}
标签：tput integer question open acm ber core and The
原文地址：https://www.cnblogs.com/ClearMoonlight/p/10529584.html
