Codeforces 797E Array Queries

题目

a is an array of n positive integers, all of which are not greater than n.

You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.

输入

The first line contains one integer n (1 ≤ n ≤ 100000).

The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n).

The third line containts one integer q (1 ≤ q ≤ 100000).

Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).

输出

Print q integers, ith integer must be equal to the answer to ith query.

样例

input
3
1 1 1
3
1 1
2 1
3 1
output
2
1
1

注意事项

Consider first example:

In first query after first operation p = 3, after second operation p = 5.

In next two queries p is greater than n after the first operation.

题解

我用了Codeforces官方Tutorial的做法.因为我觉得它的写法已经很简单易懂了，就不具体补充了。具体如下：

There are two possible solutions in O( $n^{2}$ ) time.

First of them answers each query using simple iteration — changes $p$ to
$p + a_{p} + k$ for each query until $p$ becomes greater than $n$, as stated in
the problem. But it is too slow.

Second solution precalculates answers for each $p$ and $k$:

if $p + a_{p} + k > n$, then $ans_{p,k} = 1$, else $ans_{p, k }= ans_{{p + a_{p} + k}, k }+ 1$.
But this uses O($n^2$) memory and can be done in O($n^2$) time.

Now we can notice that if $k<\sqrt{n}$, then second solution will use only $O(\sqrt{n})$time
and memory, and if $k\ge{n}$ , then first solution will do not more than $\sqrt{n}$
operations on each query. So we can combine these two solutions.

Time complexity: $O(n\sqrt{n})$

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=100000+10;
const int maxm=317;
int a[maxn],n,q,dp[maxn][maxm];
vector<int> G[maxn];

void init(void)
{
    for(int j=1;j<maxm;j++)
    {
        for(int i=n;i>=1;i--)
        {
            if(i+j+a[i]>n)
                dp[i][j]=1;
            else
                dp[i][j]=1+dp[i+j+a[i]][j];
        }
    }
}
int solve(int p,int k)
{
    int cnt=0;
    while(p<=n)
    {
        p=a[p]+p+k;
        cnt++;
    }
    return cnt;
}
int main(void)
{
    int p,k;
    cin>>n;
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    init();//precalculation:dp
    cin>>q;
    while(q--)
    {
        scanf("%d%d",&p,&k);
        if(k>315)//use brute force
            printf("%d\n",solve(p,k));
        else
            printf("%d\n",dp[p][k]);
    }

    return 0;
}