hdu2227---Find the nondecreasing subsequences(dp+树状数组)

Problem Description

How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, …., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next line contains n integers {s1, s2, s3, …., sn}, 1 <= n <= 100000, 0 <= si <= 2^31.

Output

For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007.

Sample Input

3

1 2 3

Sample Output

7

Author

8600

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dp[i]表示以第i个元素结尾的不下降序列个数

dp[i]=∑i?1j=1 dp[j] +1

n达10万,所以用树状数组来优化

/*************************************************************************
    > File Name: hdu2227.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年06月02日 星期二 18时33分24秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

static const int N = 100010;
static const int mod = 1000000007;
LL tree[N];

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

void add(int n, int x, LL val) {
    for (int i = x; i <= n; i += lowbit(i)) {
        tree[i] += val;
        tree[i] %= mod;
    }
}

LL sum(int x) {
    LL ans = 0;
    for (int i = x; i; i -= lowbit(i)) {
        ans += tree[i];
        ans %= mod;
    }
    return ans;
}

LL dp[N];
int Arr[N];
int xis[N];
int cnt;

int search(int val) {
    int l = 1, r = cnt;
    int mid;
    while (l <= r) {
        mid = (l + r) >> 1;
        if (xis[mid] == val) {
            break;
        }
        if (xis[mid] > val) {
            r = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }
    return mid;
}

int main() {
    int n;
    while (~scanf("%d", &n)) {
        cnt = 0;
        memset(tree, 0, sizeof(tree));
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &Arr[i]);
            xis[++cnt] = Arr[i];
        }
        sort(xis + 1, xis + cnt + 1);
        cnt = unique(xis + 1, xis + cnt + 1) - xis - 1;
        memset(dp, 0, sizeof(dp));
        LL ans = 0;
        for (int i = 1; i <= n; ++i) {
            int val = search(Arr[i]);
            LL Sum = sum(val);
            dp[i] = 1 + Sum;
            dp[i] %= mod;
            add(n, val, dp[i]);
            ans += dp[i];
            ans %= mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}