HDU 2227 Find the nondecreasing subsequences(DP)

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

题意：问你不降子序列的个数。

一看n达到了1e5的级别。就知道得nlogn算法。

然而想到了一个mp的迭代可是每次迭代都得log复杂度太高。所以树状数组+离散化搞。

题解：设dp[i]为前i个而且以i结尾的的不降子序列个数。

我们知道前面凡是小于等于a[i]的都能够到dp[i],所以dp[i]+=dp[j](a[j]<=a[i]&&j<i).

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<vector>

#include<string>

#include<iostream>

#include<queue>

#include<cmath>

#include<map>

#include<stack>

#include<bitset>

using namespace std;

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

#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )

#define CLEAR( a , x ) memset ( a , x , sizeof a )

typedef long long LL;

typedef pair<int,int>pil;

const int INF = 0x3f3f3f3f;

const int MOD=1e9+7;

const int maxn=1e5+10;

int a[maxn],n,b[maxn+100];

LL dp[maxn],c[maxn+100];

int lowbit(int x)

{

    return x&(-x);

}

void update(int x,LL val)

{

    while(x<maxn)

    {

        c[x]=(c[x]+val)%MOD;

        x+=lowbit(x);

    }

}

LL query(int x)

{

    LL sum=0;

    while(x>0)

    {

        sum=(sum+c[x])%MOD;

        x-=lowbit(x);

    }

    return sum;

}

int main()

{

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

    {

        int cnt=1;

        CLEAR(c,0);

        REPF(i,1,n)

        {

           scanf("%d",&a[i]);

           b[cnt++]=a[i];

        }

        sort(b+1,b+cnt);

        cnt=unique(b+1,b+cnt)-(b+1);

        for(int i=1;i<=n;i++)

            dp[i]=1;

        LL ans=0;

        for(int i=1;i<=n;i++)

        {

            int x=lower_bound(b+1,b+1+cnt,a[i])-b;

            dp[i]=(dp[i]+query(x))%MOD;

            update(x,dp[i]);

            ans=(ans+dp[i])%MOD;

        }

        printf("%I64d\n",ans);

    }

    return 0;

}

/*

*/

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HDU 2227 Find the nondecreasing subsequences(DP)

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