Codeforces Round #174 (Div. 1) B. Cow Program

B. 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:

Initially, x = 1 and y = 0. If, after any step, x ≤ 0 or x > 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 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.

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

and y becomes 1 + 2 = 3.

For i = 2,  x becomes

and y becomes 2 + 4 = 6.

For i = 3,  x becomes

and y becomes 3 + 1 + 4 = 8.

思路： 记忆化搜索

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <iostream>
#define LL long long
#define maxn 200010
#define INF 0x3f3f3f3f
using namespace std;

int a[maxn] ,n ;
LL dp[maxn][2];
bool vi[maxn] ;

LL dfs(int x,int v)
{
// cout<<x<<endl;
if(x<=0||x>n) return 0 ;
if(dp[x][v] != -1)
{
return dp[x][v] ;
}
dp[x][v]=-2;
if(v==0){
LL ans=dfs(a[x]+x,1-v) ;
if(ans==-2) return ans;
return dp[x][v]=ans+a[x];
}
else {
LL ans=dfs(x-a[x],1-v) ;
if(ans==-2) return  ans;
return dp[x][v]=ans+a[x];
}
}
int main()
{
int i,j;
LL ans;
int T ,m1,m2;
while(scanf("%d",&n) != EOF)
{
for(i = 2 ; i <= n ;i++)
scanf("%d",&a[i]) ;
memset(dp,-1,sizeof(dp)) ;
for(i = 1 ; i < n ;i++)
{
a[1]=i;
// puts("==");
ans=dfs(1+a[1],1);
if(ans==-2)ans=-1;
else ans+=a[1];
printf("%I64d\n",ans);
}
}
return 0;
}