【题目链接】:http://codeforces.com/problemset/problem/314/C

【题意】



让你从n个元素的数组中选出所有的不同的非递减子数列;

然后计算比这个子数列小的和它的长度一样长的数列的个数;

“小”的定义在题目里有说;

【题解】



设dp[i]表示以i作为非递减子数列的最后一个数的比它小的数列的个数;

则有递推式

dp[i] = (dp[1]+dp[2]+…+dp[i])*i+i;

写个树状数组,来快速求和就好;

要写出原数组,维护原数组;

不然求dp[i]比较麻烦;

最后输出∑dpi就好



【Number Of WA】



1(数组开小了)



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

#define Open() freopen("F:\\rush.txt","r",stdin)

#define Close() ios::sync_with_stdio(0),cin.tie(0)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 1e6;

const int N1 = 1e5;

const int MOD = 1e9+7;

struct BIT{

    LL a[N+10];

    int lowbit(int x){

        return x&(-x);

    }

    void add(int x,LL y){

        while (x<=N){

            a[x] = (a[x]+y)%MOD;

            x+=lowbit(x);

        }

    }

    LL sum(int x){

        LL temp = 0;

        while (x>0){

            temp = (temp+a[x])%MOD;

            x-=lowbit(x);

        }

        return temp;

    }

}c;

int n;

LL dp[N+10];

int main(){

    //Open();

    Close();

    cin >> n;

    rep1(i,1,n){

        int x;

        cin >> x;

        LL temp = (c.sum(x)*x + x)%MOD;

        c.add(x,(temp-dp[x]+MOD)%MOD);

        dp[x] = temp;

    }

    cout << c.sum(N) << endl;

    return 0;

}