zoj 3929 Deque and Balls(需要思考的递推)

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cstdlib>
 4 
 5 #include<string>
 6 #include<algorithm>
 7 #include<set>
 8 #include<map>
 9 #include<vector>
10 #include<stack>
11 #include<queue>
12 #include<cmath>
13 
14 #include<iostream>
15 using namespace std;
16 
17 typedef long long ll;
18 const int mod = 1000000007;
19 ll ans[100005],two[100005],num[100005],sum[100005];
20 
21 /*
22     num[a] : 用来保存已经出现 a 需要减去的次数 
23     num[tmp] = (num[tmp] + two[i-1]) % mod;  越近相邻次数越多,如果两数相同,减去的也越多
24     
25      *sum[i] = (sum[i-1] + two[i]) % mod;
26                 情况和    
27                 
28     ans[i] = (2*ans[i-1] + sum[i - 2] - num[tmp] + mod) % mod;
29                  之前的情况和  加上这个的 
30               两端都可插入  
31                  之前情况*2 
32 */
33 
34 void init(){
35     two[0] = two[1] = 1;
36     for(int i = 2 ; i < 100005 ; i ++){
37         two[i] = (2*two[i-1])%mod;
38     }
39     sum[0] = 1;
40     for(int i = 1 ; i < 100005 ; i ++){
41         sum[i] = (sum[i-1] + two[i]) % mod;
42     }
43     
44 }
45 
46 int main(){
47     
48     int T;
49     init();
50     scanf("%d",&T);
51     while(T--){
52         int n;
53         scanf("%d",&n);
54         memset(ans,0,sizeof(ans));
55         memset(num,0,sizeof(num)); 
56         for(int i = 1 ,tmp ; i <= n ; i ++){
57             scanf("%d",&tmp);
58             ans[i] = (2*ans[i-1] + sum[i - 2] - num[tmp] + mod) % mod;
59             num[tmp] = (num[tmp] + two[i-1]) % mod;
60         }
61         printf("%lld\n",(ans[n]*2)%mod);
62     }
63     
64     return 0;
65 }