1202 number of sub-sequences

1202 sub-sequence number base time limit: 1 second space limit: 131072 KB sub-sequence definition: For a sequence a=a[1],a[2],...... a[n]. The non-empty sequence a ' =a[p1],a[p2]......a[pm] is a sub-sequence of a, where 1<=p1<p2<.....<pm<=n. For example, 4,14,2,3 and 14,1,2,3 are all sub-sequences of 4,13,14,1,2,3. For the given sequence a, some sub-sequences may be the same, only 1 are counted here, please output a number of different sub-sequences. Because the answer is large, output mod 10^9 + 7 results. Input

Line 1th: A number N, indicating the length of the sequence (1 <= N <= 100000) 2-n + 1 lines: Elements in the sequence (1 <= a[i] <= 100000)

Output

The number of different sub-sequences of output a mod 10^9 + 7.

Input example

41232

Output example

13
Idea: Dp.dp[i] represents the first I can be composed of a number of different sub-sequences, then consider whether the current value in the previous occurrence, did not appear is dp[i]=2*dp[i-1]+1, appear for 2*dp[i-1]-Dp[id[ans[i]]-1] The last bit is the same as the previous one and the current value.

1#include <stdio.h>2#include <algorithm>3#include <iostream>4#include <stdlib.h>5#include <queue>6#include <string.h>7#include <Set>8#include <map>9#include <math.h>Ten using namespacestd; One intans[1000005]; A Const intMoD = 1e9+7; -typedefLong LongLL; -LL dp[1000005]; the intid[1000005]; -map<int,int>my; - intMainvoid) - { +         intN; -          while(SCANF ("%d", &n)! =EOF) +         { A                  for(inti =1; I <= N; i++) at                 { -scanf"%d",&ans[i]); -                 } -Memset (DP,0,sizeof(DP)); -memset (ID,0,sizeof(ID)); -                  for(inti =1; I <= N; i++) in                 { -                         if(!Id[ans[i]]) to                         { +Dp[i] =2*dp[i-1]%mod +1; -Dp[i]%=MoD; theId[ans[i]] =i; *                         } $                         ElsePanax Notoginseng                         { -Dp[i] = (2*dp[i-1]%mod-dp[id[ans[i]]-1])%mod+MoD; thedp[i]%=MoD; +Id[ans[i]] =i; A                         } the                 } +printf"%lld\n", Dp[n]); -         } $         return 0; $}

1202 number of sub-sequences