HUST 1601 Shepherd

Description

Hehe keeps a flock of sheep, numbered from 1 to n and each with a weight 

wi. To keep the sheep healthy, he prepared some training for his sheep. 

Everytime he selects a pair of numbers (a,b), and chooses the sheep with number a, a+b, a+ 

2b, … to get trained. For the distance between the sheepfold and the training site is too far, he needs to arrange a truck with appropriate loading capability to transport those sheep. So he wants to know the total weight of the sheep he selected each time, and he finds you to help him. 

Input

There’re several test cases. For each case: 

The first line contains a positive integer n ( 

1≤n≤10^5)---the number of sheep 

Hehe keeps. 

The second line contains n positive integer 

wi( 

1≤n≤10^9), separated by spaces, where the 

i-th number describes the weight of the 

i-thsheep. 

The third line contains a positive integer q ( 

1≤q≤10^5)---the number of training plans 

Hehe prepared. 

Each following line contains integer parameters a and b ( 

1≤a, 

b≤n)of the corresponding plan. 

Output

For each plan (the same order in the input), print the total weight of sheep selected. 

Sample Input

5

1 2 3 4 5

3

1 1

2 2

3 3

Sample Output

15

6

3

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
int n,m,a[maxn],x,y;
LL sum[maxn][61];

int main()
{
  while (~scanf("%d",&n))
  {
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);
    int k=min(60,n);
    for (int i=n;i>=1;i--)
    {
      for (int j=1;j<=k;j++)
      {
        if (i+j<=n) sum[i][j]=sum[i+j][j]+a[i];
        else sum[i][j]=a[i];
      }
    }
    scanf("%d",&m);
    while (m--)
    {
      scanf("%d%d",&x,&y);
      if (y>k)
      {
        LL ans=0;
        for (int i=x;i<=n;i+=y) ans+=a[i];
        printf("%lld\n",ans);
      }
      else if (y) printf("%lld\n",sum[x][y]);
      else printf("%d\n",a[x]);
    }
  }
  return 0;
}