1002(dp+树状数组+离散化)

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

题目大概：

求一个序列的子序列数。

思路：

用dp推出答案，用树状数组优化。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
int n;
int b[100010];
long long c[100010];
int dp[100010];
int pp=1000000007;
struct poin
{
    int v,id;
}a[100010];

bool cmp(const poin a,const poin b)
{
    return a.v<b.v;
}


int lowbit(int x)
{
    return x&(-x);
}
int add(int x,int v)
{
    while(x<=100001)
    {
        c[x]+=v;
        c[x]=c[x]%pp;
        x=x+lowbit(x);
    }
}

long long sum(int x)
{
    long long su=0;
    while(x>0)
    {
        su+=c[x];
        su=su%pp;
        x-=lowbit(x);
    }
    return su;
}
int main()
{





        while(~scanf("%d",&n))
        {

         memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].v);
            a[i].id=i;
        }

        sort(a+1,a+n+1,cmp);
        b[a[1].id]=1;
        int o=2;
        for(int i=2;i<=n;i++)
        {
            if(a[i].v==a[i-1].v)b[a[i].id]=b[a[i-1].id];
            else b[a[i].id]=o++;
        }
       for(int i=1;i<=n;i++)
       {
           dp[i]=sum(b[i])+1;
           add(b[i],dp[i]);
       }



            printf("%d\n",sum(n));


        }

    return 0;
}