Codeforces 284D Cow Program【思维+记忆化搜索】

D. Cow Program

time 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 with! The program contains two integer variables, x and y, and performs the following operations on a sequence a1, a2, ..., an of positive integers:

1. Initially, x = 1 and y = 0. If, after any step, x ≤ 0 or x > n, the program immediately terminates.
2. The program increases both x and y by a value equal to ax simultaneously.
3. The program now increases y by ax while decreasing x by ax.
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 sequence a2, a3, ..., an. Suppose for each i (1 ≤ i ≤ n - 1) we run the program on the sequence i, a2, a3, ..., an. For each such run output the final value of y if the program terminates or -1 if it does not terminate.

Input

The first line contains a single integer, n (2 ≤ n ≤ 2·105). The next line contains n - 1 space separated integers, a2, a3, ..., an(1 ≤ ai ≤ 109).

Output

Output n - 1 lines. On the i-th line, print the requested value when the program is run on the sequence i, a2, a3, ...an.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Examples

input

4
2 4 1

output

3
6
8

input

3
1 2

output

-1
-1

Note

In the first sample

1. For i = 1,  x becomes  and y becomes 1 + 2 = 3.
2. For i = 2,  x becomes  and y becomes 2 + 4 = 6.
3. For i = 3,  x becomes  and y becomes 3 + 1 + 4 = 8.

题目大意：

给出a【2】~a【n】，让我们给a【1】赋值1~n-1，从而计算结果。

一开始x=1，y=0，然后对于一个回合：

x+=a【x】，y+=a【x】；

x-=a【x】，y+=a【x】；

过程中如果x<=0||x>n为终止条件，问终止的时候y的价值，如果能够无限循环下去，输出-1.

思路：

考虑问题的本质，当我们此时模拟到点x的时候，无非接下来有两种操作：

①x+=a【x】；

②x-=a【x】；

那么如果我们到达x点进行过了操作①而后又到达x点需要进行操作①的时候，我们肯定就知道接下来会有循环存在了。

那么设定vis【x】【2】表示到点x应该进行的操作为①或者②的情况是否发生过。

那么过程相当于一个记忆化搜索。

维护一下就行了。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
ll n;
ll a[250000];
ll dp[250000][2];
int vis[250000][2];
void Dfs(ll i,ll j)
{
    if(vis[i][j]==0)
    {
        vis[i][j]=1;
        ll nex;
        if(j==1)nex=i-a[i];
        else nex=i+a[i];
        if(nex<=0||nex>n)
        {
            dp[i][j]=a[i];
        }
        else
        {
            Dfs(nex,j^1);
            if(dp[nex][j^1]!=-1)dp[i][j]=dp[nex][j^1]+a[i];
            else return ;
        }
    }
    else return ;
}
int main()
{
    while(~scanf("%I64d",&n))
    {
        memset(vis,0,sizeof(vis));
        memset(dp,-1,sizeof(dp));
        vis[1][0]=1;
        vis[1][1]=1;
        dp[1][1]=0;
        for(ll i=2;i<=n;i++)scanf("%I64d",&a[i]);
        for(ll i=2;i<=n;i++)Dfs(i,1);
        for(ll i=2;i<=n;i++)
        {
            if(dp[i][1]!=-1)dp[i][1]+=i-1;
        }
        for(ll i=2;i<=n;i++)printf("%I64d\n",dp[i][1]);
    }
}