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。
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 }