HDOJ 题目2227 Find the nondecreasing subsequences（树状数组，离散化，DP）

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Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1728 Accepted Submission(s): 631

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

Author

8600

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ac代码

15352698

2015-11-02 18:35:20

Accepted

2227

1123MS

7632K

1190 B

G++

XY_

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<map>
#define INF 0x3f3f3f3f
#define LL long long
#define mod 1000000007
using namespace std;
int val[100010],t[100010];
int a[100010],cnt;
int lowbit(int x)
{
    return x&(-x);
}
void add(int p,int val)
{
    while(p<=cnt)
    {
        a[p]=(a[p]+val)%mod;
        p+=lowbit(p);
    }
}
int getsum(int x)
{
    int ans=0;
    while(x>0)
    {
        ans=(ans+a[x])%mod;
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int i;
        memset(a,0,sizeof(a));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&val[i]);
            t[i]=val[i];
        }
        sort(t+1,t+n+1);
        cnt=unique(t+1,t+1+n)-(t+1);
        map<int ,int>mp;
        for(i=1;i<=cnt;i++)
        {
            mp[t[i]]=i;
        }
        int ans=0;
        add(1,1);
        for(i=1;i<=n;i++)
        {
            int x=getsum(mp[val[i]]);
           // printf("%d\n",x);
            ans=(ans+x)%mod   ;
            add(mp[val[i]],x);
        }
        printf("%d\n",ans);
    }
}