HDU 2227 Find the nondecreasing subsequences(DP)

全栈程序员站长

发布于 2022-07-06 11:42:49

1210

发布于 2022-07-06 11:42:49

文章被收录于专栏：全栈程序员必看

大家好，又见面了，我是全栈君。

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;
}

/*

*/

发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/117125.html原文链接：https://javaforall.cn

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2022年1月1，如有侵权请联系 cloudcommunity@tencent.com 删除

java