CF 174 (div2) D

D. Cow Programtime limit per test

2 seconds

Memory limit per test

256 megabytes

Input

Standard input

Output

Standard output

Farmer John has just given the cows a program to play! The program contains two integer variables,XAnd
Y, And performs the following operations on a sequenceA_{1, bytes,A2, middle..., middle ,...,AN
Of positive integers:}

1. Initially,XBetween = between 1 and
   YWhen = then 0. If, after any step,XLimit ≤ limit 0 orXLatency> latencyN, The program immediately terminates.
2. The program increases bothXAnd
   YBy a value equalA_{XSimultaneously.}
3. The program now increasesYBy
   A_{XWhile decreasing
   XByAX.}
4. The program executes steps 2 and 3 (first step 2, then step 3) repeatedly until it terminates (it may never terminate ). so, the sequence of executed steps may start with: step 2, step 3, step 2, step 3, step 2 and so on.

The cows are not very good at arithmetic though, and they want to see how the program works. Please help them!

You are given the sequenceA_{2, bytes,A3, middle..., middle ,...,AN. Suppose for eachI
(1 digit ≤ DigitILimit ≤ limitNUpload-sequence 1) we run the program on the sequenceI, Bytes,A2, bytes,A3, middle..., middle ,...,AN.
For each such run output the final valueYIf the program terminates or
-1 if it does not terminate.}

Input

The first line contains a single integer,N(2 cores ≤ CoresNLimit ≤ limit 2 · 10^{5). The next line containsNEncryption-rule 1 space separated integers,
A_{2, bytes,A3, middle..., middle ,...,AN(1 digit ≤ DigitAILimit ≤ limit 109 ).}}

Output

OutputNLimits-limits 1 lines. On
I-Th line, print the requested value when the program is run on the sequenceI, Bytes,A_{2, bytes,A3, large ,...AN.}

Please do not use the % lld specifier to read or write 64-bit integers in bytes ++. It is preferred to use thecin,
Cout streams or the % I64d specifier.

Sample test (s) Input

42 4 1

Output

368

Input

31 2

Output

-1-1

Note

In the first sample

1. ForIPriority = Priority 1, priority,XBecomes andY
   Becomes 1 response + response 2 response = Response 3.
2. ForILimit = Limit 2, limit,XBecomes andY
   Becomes 2 centers + limit 4 centers = Limit 6.
3. ForILatency = latency 3, latency,XBecomes andY
   Becomes 3 cores + cores 1 cores + cores 4 cores = cores 8.

If you do this directly, you need to use the memory to search for the cumulative increment of y until the end of each x. Note that the memory status here is two-dimensional, because operation 1 is performed when the result reaches x, or 2 has different effects on the results!

# Include <cstdio> # include <cstring> # include <iostream> # include <algorithm> # include <vector> # include <stack> # include <queue> # define LL long longusing namespace std; const int maxn = 200000 + 5; int n; LL num [maxn]; int vis [maxn] [2]; LL go [maxn] [2]; LL find (int x, int kind) {if (go [x] [kind]) return go [x] [kind]; if (vis [x] [kind] = 1) {return go [x] [kind] =-1;} vis [x] [kind] = 1; if (kind = 1 & x + num [x]> N) {return go [x] [kind] = num [x];} else if (kind = 0 & x-num [x] <= 0) {return go [x] [kind] = num [x];} LL tem = find (x + (kind = 1? Num [x]:-num [x]), (kind + 1) % 2); // This is discussed in kind. 1 indicates that the current operation is 1, 0 indicates the operation 2 if (tem! =-1) return go [x] [kind] = tem + num [x]; else return go [x] [kind] =-1;} int main () {while (cin> n) {memset (vis, 0, sizeof (vis); memset (go, 0, sizeof (go); for (int I = 2; I <= n; I ++) cin> num [I]; LL x, y; for (int I = 2; I <= n; I ++) {y = I-1; LL tem = find (I, 0); if (tem =-1) {cout <-1 <endl; continue ;} y + = tem; cout <y <endl ;}} return 0 ;}