Problem Description


Let A1, A2, ... , AN be N elements. You need to deal with two kinds of operations. One type of operation is to add a given number to a few numbers in a given interval. The other is to query the value of some element.


Input


There are a lot of test cases. 
The first line contains an integer N. (1 <= N <= 50000)
The second line contains N numbers which are the initial values of A1, A2, ... , AN. (-10,000,000 <= the initial value of Ai <= 10,000,000)
The third line contains an integer Q. (1 <= Q <= 50000)
Each of the following Q lines represents an operation.
"1 a b k c" means adding c to each of Ai which satisfies a <= i <= b and (i - a) % k == 0. (1 <= a <= b <= N, 1 <= k <= 10, -1,000 <= c <= 1,000)
"2 a" means querying the value of Aa. (1 <= a <= N)


Output


For each test case, output several lines to answer all query operations.


Sample Input


4 1 1 1 1 14 2 1 2 2 2 3 2 4 1 2 3 1 2 2 1 2 2 2 3 2 4 1 1 4 2 1 2 1 2 2 2 3 2 4


Sample Output


1 1 1 1 1 3 3 1 2 3 4

1


多个树状数组

根据操作，要每次隔开操作，为了使用树状数组，所以按照k的大小和余数分成多个树状数组。

把不同的点放到同一个线上，用树状数组记录操作的情况，然后最后相加。

#include<iostream>  
#include<algorithm>
#include<math.h>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
const int maxn = 50005;
const int low(int x){ return (x&-x); }
int A[maxn], n, m, i, j, f[11][10][maxn];

void add(int x, int y, int z, int w)
{
  for (int i = z; i <= n; i += low(i)) f[x][y][i] += w;
}

long long sum(int x, int y, int z)
{
  long long total = 0;
  for (int i = z; i > 0; i -= low(i)) total += f[x][y][i];
  return total;
}

int main(){
  while (~scanf("%d", &n))
  {
    memset(A, 0, sizeof(A));
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
    cin >> m;
    while (m--)
    {
      int a, b, c, d, e;
      scanf("%d%d", &e, &a);
      if (e == 2)
      {
        long long tot = A[a];
        for (int i = 1; i <= 10; i++) tot += sum(i, a%i, (a - 1) / i + 1);
        cout << tot << endl;
      }
      else
      {
        scanf("%d%d%d", &b, &c, &d);
        add(c, a%c, (a - 1) / c + 1 , d);
        add(c, a%c, (b - (a - 1) % c - 1) / c + 2, -d);
      }
    }
  }
  return 0;
}