[Explanations] large numbers of friends

solution:

Meaning of the questions we ask is to let each with a [i] at the end of the longest sub-sequence and did not fall.

Then we first pretreatment did not fall out of a length of the longest sequence at the end of every number, in E [i], i.e. to the i-th end.

Can be obtained with a base dp.

Then we define the sum array, it represents the longest i did not fall and the end of the subsequence.

Then the number of i enumerator before, are: When e [i] == e [j] +1 time, and a [i]> = a [j], i.e., when the end i can be, sum [i] = sum [j] + a [i], can directly out.

Final output just fine.

Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define inf 2147483647
using namespace std;
int len=1;
int n,a[20001];
int sum[20001];
int pre[20001];
int e[20001];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)sum[i]=a[i],e[i]=1;
    /*for(int i=2;i<=n;i++){
        pre[1]=a[1];
        for(int j=2;j<=i;j++){
            if(a[j]>=pre[len])pre[++len]=a[j];
            else{
                int p=upper_bound(pre+1,pre+len+1,a[j])-pre;
                pre[p]=a[j];
            }
        }
        e[i]=len;
        len=1;
        memset(pre,0,sizeof(pre));
    }*/
    for(int i=2;i<=n;i++)
        for(int j=1;j<i;j++)
            if(a[i]>=a[j])e[i]=max(e[i],e[j]+1);
    for(int i=1;i<=n;i++)
        for(int j=1;j<i;j++)
            if(e[j]+1==e[i]&&a[i]>=a[j]){
                sum[i]=sum[j]+a[i];
                break;
            }
    //for(int i=1;i<n;i++)if(sum[i+1]<sum[i])sum[i+1]=sum[i];
    for(int i=1;i<=n;i++)printf("%d ",sum[i]);
    return 0;
}

Note that I use the comment section above (nlogn) algorithm for the length of time out, however, note that the reciprocal determination and in the second cycle.

[Explanations] large numbers of friends

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