HDU 2227 Find the nondecreasing subsequences （数状数组）

Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1072 Accepted Submission(s): 370

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

using namespace std;

const int mod=1000000007;
const int N=100010;

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

int n,b[N],c[N],arr[N];

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

void update(int i,int val){
    while(i<=n){
        arr[i]+=val;
        arr[i]%=mod;
        i+=lowbit(i);
    }
}

int Sum(int i){
    int ans=0;
    while(i>0){
        ans+=arr[i];
        ans%=mod;
        i-=lowbit(i);
    }
    return ans;
}

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

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d",&n)){
        memset(b,0,sizeof(b));
        memset(arr,0,sizeof(arr));
        for(int 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(int i=2;i<=n;i++)
            if(a[i].val==a[i-1].val)
                b[a[i].id]=b[a[i-1].id];
            else
                b[a[i].id]=i;
        for(int i=1;i<=n;i++){
            c[i]=Sum(b[i]);
            update(b[i],c[i]+1);
        }
        printf("%d
",Sum(n));
    }
    return 0;
}