hdu-2227-dp+bit Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2213 Accepted Submission(s): 858

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

　　　　令f(i)表示以第i个元素为结尾的非递减子序列的个数，有 f(i)=SUM{f(j) | j<i&&a[j]<=a[i]}。

用BIT来维护f，C[x]表示所有的f总和,下标反映的就是a[i]得值，这样在求解f(i)=sum(a[i])就好了。

　注意到ai范围较大，离散化处理一下。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ULL unsigned long long 
 4 #define LL long long 
 5 LL mod=1e9+7;
 6 LL C[100010];
 7 int N;
 8 struct node{
 9     int v,d;
10     bool operator<(const node& C)const{
11         if(v!=C.v) return v<C.v;
12         return d<C.d;
13     }
14 }a[101010];
15 bool cmp(node A,node B){return A.d<B.d;}
16 int main(){
17     int i,j;
18     while(scanf("%d",&N)==1){
19         LL ans=0;
20         for(i=1;i<=N;++i) scanf("%d",&a[i].v),a[i].d=i;
21         sort(a+1,a+1+N);
22         for(i=1;i<=N;++i) a[i].v=i;
23         sort(a+1,a+1+N,cmp);
24         for(i=1;i<=N;++i){
25             LL tmp=1;
26             for(int x=a[i].v;x>0;x-=(x&-x)) (tmp+=C[x])%=mod;
27             for(int x=a[i].v;x<=N;x+=(x&-x)) (C[x]+=tmp)%=mod;
28             (ans+=tmp)%=mod;
29         }
30         printf("%lld
",ans);
31         memset(C,0,sizeof(C));
32     }
33     return 0;
34 }

hdu-2227-dp+bit Find the nondecreasing subsequences

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