A Simple Problem with Integers(树状数组HDU4267)

A Simple Problem with Integers
Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4494 Accepted Submission(s): 1384

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
Source
2012 ACM/ICPC Asia Regional Changchun Online
比赛的时候没有注意到k的值很小果断超时.
更新区间(a,b)中(i-a)%k==0的点其实就是更新区间(a,b)中i%k==a%k的值,所以可以用树状数组实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 55000;

int FK[MAX][11][11];

int num[MAX];

int n,m;

int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int k,int mod,int va)//更新摸为k,取模后为mod的区间的数组
{
    while(x>0)
    {
        FK[x][k][mod]+=va;
        x-=lowbit(x);
    }
}
int Query(int x,int a)//查询a所在的区间的增加值
{
    int s=0;
    while(x<MAX)
    {
        for(int i=1;i<=10;i++)
        {
            s+=FK[x][i][a%i];
        }
        x+=lowbit(x);
    }
    return s;
}

int main()
{
    int flag,a,b,k,va;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
        }
        scanf("%d",&m);
        memset(FK,0,sizeof(FK));
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&flag);
            if(flag==1)
            {
                scanf("%d %d %d %d",&a,&b,&k,&va);
                update(b,k,a%k,va);
                update(a-1,k,a%k,-va);
            }
            else
            {
                scanf("%d",&a);
                printf("%d\n",Query(a,a)+num[a]);
            }
        }
    }
    return 0;
}