Zoj 3929 Deque and Balls (recursive to think)

Ideas for reference from: http://blog.csdn.net/doris1104/article/details/51126910 thanks

Find out the pattern of the last ball adjacent to the front ball (to satisfy the left greater than the right) and 2^ two ball spacing related

-----> The closer the number of neighbors.

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>Ten#include <stack> One#include <queue> A#include <cmath> -  -#include <iostream> the using namespacestd; -  -typedefLong Longll; - Const intMoD =1000000007; +ll ans[100005],two[100005],num[100005],sum[100005]; -  + /* A Num[a]: Used to save the number of times a needs to be subtracted at num[tmp] = (Num[tmp] + two[i-1])% MoD; the closer the number of neighbors, the more you subtract if the two numbers are the same -      - *sum[i] = (Sum[i-1] + two[i])% MoD; - Situation and -                  - Ans[i] = (2*ans[i-1] + sum[i-2]-num[tmp] + MoD)% MoD; in The previous situation and the addition of this - Both ends can be inserted to Previous conditions + */ -  the voidinit () { *two[0] = two[1] =1; $      for(inti =2; I <100005; i + +){Panax NotoginsengTwo[i] = (2*two[i-1])%MoD; -     } thesum[0] =1; +      for(inti =1; I <100005; i + +){ ASum[i] = (sum[i-1] + two[i])%MoD; the     } +      - } $  $ intMain () { -      -     intT; the init (); -scanf"%d",&T);Wuyi      while(t--){ the         intN; -scanf"%d",&n); Wumemset (ans,0,sizeof(ans)); -memset (NUM,0,sizeof(num));  About          for(inti =1, TMP; I <= N; i + +){ $scanf"%d",&tmp); -Ans[i] = (2*ans[i-1] + sum[i-2]-num[tmp] + MoD)%MoD; -NUM[TMP] = (Num[tmp] + two[i-1]) %MoD; -         } Aprintf"%lld\n", (ans[n]*2)%MoD); +     } the      -     return 0; $}

Zoj 3929 Deque and Balls (recursive to think)