Codeforces Beta Round #80 (Div. 1 Only) D. Time to Raid Cowavans 分块

D. Turtles

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/103/problem/D

Description

As you know, the most intelligent beings on the Earth are, of course, cows. This conclusion was reached long ago by the Martian aliens, as well as a number of other intelligent civilizations from outer space.

Sometimes cows gather into cowavans. This seems to be seasonal. But at this time the cows become passive and react poorly to external stimuli. A cowavan is a perfect target for the Martian scientific saucer, it's time for large-scale abductions, or, as the Martians say, raids. Simply put, a cowavan is a set of cows in a row.

If we number all cows in the cowavan with positive integers from 1 to n, then we can formalize the popular model of abduction, known as the (a, b)-Cowavan Raid: first they steal a cow number a, then number a + b, then — number a + 2·b, and so on, until the number of an abducted cow exceeds n. During one raid the cows are not renumbered.

The aliens would be happy to place all the cows on board of their hospitable ship, but unfortunately, the amount of cargo space is very, very limited. The researchers, knowing the mass of each cow in the cowavan, made p scenarios of the (a, b)-raid. Now they want to identify the following thing for each scenario individually: what total mass of pure beef will get on board of the ship. All the scenarios are independent, in the process of performing the calculations the cows are not being stolen.

Input

The first line contains the only positive integer n (1 ≤ n ≤ 3·10⁵) — the number of cows in the cowavan.

The second number contains n positive integer w_i, separated by spaces, where the i-th number describes the mass of the i-th cow in the cowavan (1 ≤ w_i ≤ 10⁹).

The third line contains the only positive integer p — the number of scenarios of (a, b)-raids (1 ≤ p ≤ 3·10⁵).

Each following line contains integer parameters a and b of the corresponding scenario (1 ≤ a, b ≤ n).

 

Output

Print for each scenario of the (a, b)-raid the total mass of cows, that can be stolen using only this scenario.

 

Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams of the %I64d specificator.

 

Sample Input

3
1 2 3
2
1 1
1 2

Sample Output

6
4

HINT

 

题意

给你N个数，然后每次询问给你(x,y)，求a[x]+a[x+y]+a[x+2*y]+a[x+3*y]的和是多少

 

题解：

分块做，把cy[i]分成sqrt(n)，把小于的都预存起来，大于的就直接暴力就好

小的，我们就可以利用 dp的思想搞定

然后复杂度O(sqrt(n))

详情见：2014年国家集训队论文《根号算法，不只是分块》

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 400001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

ll a[maxn],n,m;
int cx[maxn],cy[maxn];
ll ans[maxn];
ll dp[maxn];
int kiss=600;
vector<int> d[600];
int main()
{
    //test;
    n=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    int m=read();
    for(int i=1;i<=m;i++)
    {
        cx[i]=read(),cy[i]=read();
        if(cy[i]<kiss)
            d[cy[i]].push_back(i);
        else
        {
            ll ans1=0;
            for(int j=cx[i];j<=n;j+=cy[i])
            {
                ans1+=a[j];
            }
            ans[i]=ans1;
        }
    }
    for(int i=1;i<kiss;i++)
    {
        if(d[i].size())
        {
            for(int j=n;j;j--)
            {
                if(j+i>n)
                    dp[j]=a[j];
                else
                    dp[j]=dp[j+i]+a[j];
            }
            for(int j=0;j<d[i].size();j++)
                ans[d[i][j]]=dp[cx[d[i][j]]];
        }
    }
    for(int i=1;i<=m;i++)
        cout<<ans[i]<<endl;
}