codeforces 314C Sereja and Subsequences（树状数组+DP）

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 modulo1000000007 (10^9 + 7).

Input

The first line contains integer n (1 ≤ n ≤ 10^5). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 10^6).

Output

In the single line print the answer to the problem modulo 1000000007 (10^9 + 7).

题目大意：

　　给定一个数组a：

　　（1）定义某个序列a小于某个序列b当且仅当a[i]≤b[i]对所有i都成立；（2）写出a的不同的所有非下降子序列；（3）若每个不同的非下降子序列的所有小于它的序列数之和定义为f，求所有f的总和。

思路：

　　若用s[i]表示以a[i]结尾的所有非下降子序列的f之和，那么设a[j]≤a[i]且j<i，那么s[i] = sum(s[j])*a[i] + a[i]。即在序列a[j]后接上一个a[i]可形成新的子序列。

　　要维护一段数字的和，树状数组就派上用场了。

　　这里还有一个问题，就是有可能多个a[i]相同，这样会导致重复计算。我们只需要把s[i]减去a[j]≤a[i]且j<i的sum(s[j])即可。

　　比如要计算序列2 2 2的时候，s[1] = {a[1]}，s[2] = {a[1,2]}，s[3] = {a[1,2,3]}。此时s[i]准确表示为：以a[i]结尾的与i之前所有子序列都不同的所有非下降子序列的f之和。

#include <cstdio>
#include <algorithm>
using namespace std;

#define MOD 1000000007
#define MAXN 100010

struct node{
    int x, rank, pos;
} a[MAXN];

bool cmpx(node a, node b){
    return a.x < b.x;
}

bool cmppos(node a, node b){
    return a.pos < b.pos;
}

int n, tot;
int b[MAXN], c[MAXN];

inline int lowbit(int x){
    return x&-x;
}

inline int sum(int x){
    int ans = 0;
    while(x > 0){
        ans = (ans + c[x])%MOD;
        x -= lowbit(x);
    }
    return ans;
}

inline void add(int i, int x){
    while(i <= tot){
        c[i] = (c[i] + x)%MOD;
        i += lowbit(i);
    }
}

int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i].x), a[i].pos = i;
    sort(a+1, a+n+1, cmpx);
    tot = 1; a[1].rank = 1;
    for(int i = 2; i <= n; ++i){
        if(a[i].x != a[i-1].x) ++tot;
        a[i].rank = tot;
    }
    sort(a+1, a+n+1, cmppos);
    for(int i = 1; i <= n; ++i){
        int tmp = ((long long)sum(a[i].rank) * a[i].x % MOD + a[i].x)%MOD;
        add(a[i].rank, (tmp - b[a[i].rank] + MOD)%MOD);
        b[a[i].rank] = tmp;
    }
    printf("%d\n",sum(tot));
}