FZU 2129 子序列个数 (递推dp)

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2129

dp[i]表示前i个数的子序列个数

当a[i]在i以前出现过，dp[i] = dp[i - 1]*2 - dp[pre - 1]，pre表示a[i]在i之前的位置

当a[i]在i以前没有出现过，dp[i] = dp[i - 1] *2 + 1

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long LL;
 typedef pair <int, int> P;
 const int N = 1e6 + ;
 LL dp[N], mod = 1e9 + , pre[N];

 int main()
 {
     int n, num;
     while(~scanf("%d", &n)) {
         memset(pre, , sizeof(pre));
         for(int i = ; i <= n; ++i) {
             scanf("%d", &num);
             if(pre[num]) {
                 dp[i] = ((dp[i - ]* - dp[pre[num] - ]) % mod + mod) % mod;
             } else {
                 dp[i] = (dp[i - ]* + ) % mod;
             }
             pre[num] = i;
         }
         printf("%lld\n", dp[n]);
     }
     return ;
 }