hdu 2227 Find the nondecreasing subsequences（DP+树状数组+离散化）

Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 503    Accepted Submission(s): 192

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

         

          题目大意：给你一个串，求这个串中不递减的子串有多少个？初一看，完全想不到会是用树状数组，但是他就是这么神奇，不递减就想到逆序数，逆序数又想到了树状数组，居然是用他。他是求子串有多少个，又要用到DP，这里DP就是用树状数组慢慢推上去，最后是注意是求和会溢出，记得%1000000007。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2227

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <algorithm>
using namespace std;

struct node
{
    int val, id;
}a[100005];

bool cmp(node a, node b)
{
    return a.val < b.val;
}

int b[100005], c[100005], s[100005], n;

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

void update(int i, int x)
{
    while(i <= n)
    {
        s[i] += x;
        if(s[i] >= 1000000007)
            s[i] %= 1000000007;
        i += lowbit(i);
    }
}

int sum(int i)
{
    int sum = 0;
    while(i > 0)
    {
        sum += s[i];
        if(sum >= 1000000007)
            sum %= 1000000007;
        i -= lowbit(i);
    }
    return sum;
}

int main()
{
    int i, res;
    while(scanf("%d", &n) != EOF)
    {
        memset(b, 0, sizeof(b));
        memset(s, 0, sizeof(s));
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &a[i].val);
            a[i].id = i;
        }
        sort(a+1, a+n+1, cmp);
        b[a[1].id] = 1;
        for(i = 2; i <= n; i++)
        {
            if(a[i].val != a[i-1].val)
                b[a[i].id] = i;
            else b[a[i].id] = b[a[i-1].id];
        }
        res = 0;
        for(i = 1; i <= n; i++)
        {
            c[i] = sum(b[i]);
            update(b[i], c[i]+1);
        }
        printf("%d\n", sum(n));
    }

    return 0;
}