CF 174(div2) D

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 sequencea1, a2, ..., an
of positive integers:

Initially, x = 1 and
y = 0. If, after any step, x ≤ 0 orx > n, the program immediately terminates.
The program increases both x and
y by a value equal to ax simultaneously.
The program now increases y by
ax while decreasing
x by ax.
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 eachi
(1 ≤ i ≤ n - 1) we run the program on the sequencei, a2, a3, ..., an.
For each such run output the final value ofy 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 containsn - 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 sequencei, a2, a3, ...an.

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

Sample test(s)

Input

4
2 4 1

Output

3
6
8

Input

3
1 2

Output

-1
-1

Note
In the first sample

For i = 1,  x becomes

andy
becomes 1 + 2 = 3.
For i = 2,  x becomes

andy
becomes 2 + 4 = 6.
For i = 3,  x becomes

andy
becomes 3 + 1 + 4 = 8.

直接做肯定超时，需要用记忆化搜索到达每个x对应的直到结束时y的累积增量，注意这里的记忆状态有二维，因为到达x时是进行操作1，还是2对结果的影响是不同的！

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#define LL long long
using 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);//分kind讨论，1代表当前为操作1，0代表操作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;
}