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

7

 

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。

 

　　

 #include<bits/stdc++.h>
 using namespace std;
 #define ULL unsigned long long
 #define LL long long
 LL mod=1e9+;
 LL C[];
 int N;
 struct node{
     int v,d;
     bool operator<(const node& C)const{
         if(v!=C.v) return v<C.v;
         return d<C.d;
     }
 }a[];
 bool cmp(node A,node B){return A.d<B.d;}
 int main(){
     int i,j;
     while(scanf("%d",&N)==){
         LL ans=;
         for(i=;i<=N;++i) scanf("%d",&a[i].v),a[i].d=i;
         sort(a+,a++N);
         for(i=;i<=N;++i) a[i].v=i;
         sort(a+,a++N,cmp);
         for(i=;i<=N;++i){
             LL tmp=;
             for(int x=a[i].v;x>;x-=(x&-x)) (tmp+=C[x])%=mod;
             for(int x=a[i].v;x<=N;x+=(x&-x)) (C[x]+=tmp)%=mod;
             (ans+=tmp)%=mod;
         }
         printf("%lld\n",ans);
         memset(C,,sizeof(C));
     }
     return ;
 }