動態規劃寫法:n個數,1--n對於每個數i判斷左右區間連續且不小於i的個數numb,即對於i,dp[i] = a[i]*numb,然後遍歷dp,找出最大值,判斷i左右區間連續且不小於i的numb時,可以開兩個數組,L和R,注意不要嵌套兩重循環,否則後超時下面是AC代碼:
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 100000+10;
__int64 dp[N], a[N];
int l[N], r[N];
int main()
{
int n;
while(~scanf("%d", &n), n)
{
for(int i=0; i<n; ++i)
{
scanf("%I64d", &a[i]);
l[i] = r[i] = i;
}
for(int i=1; i<n; ++i)
{
while(l[i]>0 && a[l[i]-1] >= a[i])
l[i] = l[l[i]-1];
}
for(int i=n-1; i>=0; --i)
{
while(r[i]<n-1 && a[r[i]+1] >= a[i])
r[i] = r[r[i]+1];
}
__int64 Max = -1;
for(int i=0; i<n; ++i)
{
dp[i] = a[i] * (r[i] - l[i] + 1);
Max = Max>dp[i] ? Max : dp[i];
}
printf("%I64d\n", Max);
}
return 0;
}
嵌套循環超時代碼:
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 100000 + 10;
int a[N], dp[N], val[N];
int main()
{
int n;
while(~scanf("%d", &n),n)
{
for(int i=0; i<n; ++i)
{
scanf("%d", &a[i]);
}
for(int i=0; i<n; ++i)
{
int Max = a[i];
for(int j=i-1; j>=0; --j)
{
if(a[j]>=a[i])
Max += a[i];
else
break;
}
for(int j=i+1; j<n; ++j)
{
if(a[j] >= a[i])
Max += a[i];
else
break;
}
dp[i] = Max;
}
int Max = -1;
for(int i=0; i<n; ++i)
{
Max = Max>dp[i] ? Max : dp[i];
}
cout << Max << endl;
}
return 0;
}