CodeForces - 314C Sereja and Subsequences （树状数组+dp）

时间：2022-03-21 07:35:37

Sereja has a sequence that consists of n positive integers, a1, a2, ..., an.

First Sereja took a piece of squared paper and wrote all distinct non-empty non-decreasing subsequences of sequence a. Then for each sequence written on the squared paper, Sereja wrote on a piece of lines paper all sequences that do not exceed it.

A sequence of positive integers x = x1, x2, ..., xr doesn't exceed a sequence of positive integers y = y1, y2, ..., yr, if the following inequation holds: x1 ≤ y1, x2 ≤ y2, ..., xr ≤ yr.

Now Sereja wonders, how many sequences are written on the lines piece of paper. Help Sereja, find the required quantity modulo 1000000007 (109 + 7).

Input

The first line contains integer n (1 ≤ n ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106).

Output

In the single line print the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input
1
42
Output
42
Input
3
1 2 2
Output
13
Input
5
1 2 3 4 5
Output
719

题意：
原题意是这样的：
看第二组样例：
3
1 2 2
首先，写出所有非空的，非严格递增的，不同的子序列：
1
2
1 2
2 2
1 2 2
然后写出小于这些子序列的数组，问数组个数
~ 1
1

~2
1 
2

~ 1 2
1 1
1 2

~2 2
1 1
1 2
2 1
2 2

~1 2 2
1 1 1
1 1 2
1 2 1
1 2 2

这样一共写出了13个数组。
显而易见的，每个子序列可以写出来的数组个数，其实就是子序列的数字之积。
思路：
dp[num[i]]表示子序列以num[i]为结尾的答案。
然后就按照输入顺序进行更新。
dp[num[i]]=（dp[1]到dp[num[i]]的和）*num[i]+num[i];
前半部分表示接在其他数字后面，用树状数组优化，后半部分表示自己单独出现
当然还要去重，就是1 2 2，第一个2会接在1后面，第二个2也接在1后面，就会重复。
用一个pre记录之前的那个dp[2],正常更新dp[2]再减去pre[2]就行了。
#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<set>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<ctime>

#define fuck(x) cout<<#x<<" = "<<x<<endl;

#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;

#define ls (t<<1)

#define rs ((t<<1)|1)

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int maxn = ;

const int maxm = ;

const int inf = 2.1e9;

const ll Inf = ;

const int mod = ;

const double eps = 1e-;

const double pi = acos(-);

int num[maxn];

ll dp[maxm];

ll a[maxm];

ll pre[maxm];

int lowbit(int x){

    return x&(-x);

}

void update(int pos,ll num){

    while (pos<maxm){

        a[pos]+=num;

        a[pos]%=mod;

        pos+=lowbit(pos);

    }

}

ll query(int pos){

    ll ans=;

    while (pos){

        ans+=a[pos];

        ans%=mod;

        pos-=lowbit(pos);

    }

    return ans;

}

int main()

{

//    ios::sync_with_stdio(false);

//    freopen("in.txt","r",stdin);

    int n;

    scanf("%d",&n);

    for(int i=;i<=n;i++){

        scanf("%d",&num[i]);

    }

    for(int i=;i<=n;i++){

        ll ans=query(num[i]);

        ll tmp=ans*num[i]+num[i];

        dp[num[i]]+=ans*num[i]+num[i];

        dp[num[i]]%=mod;

        dp[num[i]]-=pre[num[i]];

        tmp=((tmp-pre[num[i]])+mod)%mod;

        dp[num[i]]=(dp[num[i]]+mod)%mod;

        pre[num[i]]=dp[num[i]];

        update(num[i],tmp);

//        debug(dp,num[i]);

    }

    ll ans=;

    for(int i=;i<=maxm;i++){

        ans+=dp[i];

        ans%=mod;

    }

    printf("%lld\n",ans);

    return ;

}

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