A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题意：给你一个序列，有两个操作，1是将区间[a,b]内满足 (i - a) % k == 0的位置加上c，2是查询a位置的数。

分析：因为k比较小，所以可以对于每一个k建立一个树状数组，这样还不够，无法确定应该向哪个位置加数，再加一维表示模k的余数，这样建立就行了。

对于更新，我们只需要对F[k][a%k][x]的树状数组更新，因为是区间更新，可以考虑加一段和剪一段，在a这个位置更新c，在b+1的位置减去c就不影响后面了。

查询就普通查询，注意要加上初始值。

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

const int MAXN = 50000+100;
int F[12][12][MAXN];
int num[MAXN];
int n;

void update(int s,int t,int x,int val)
{
    while(x<=n)
    {
        F[s][t][x]+=val;
        x+=x&-x;
    }
}
int query(int s,int t,int x)
{
    int res=0;
    while(x>0)
    {
        res+=F[s][t][x];
        x-=x&-x;
    }
    return res;
}

int main()
{
    int m;
    int op,a,b,k,c;
    while(scanf("%d",&n)!=EOF)
    {
        memset(F,0,sizeof(F));
        for(int i=1;i<=n;i++) scanf("%d",&num[i]);
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d%d%d%d",&a,&b,&k,&c);
                int t=a%k;
                update(k,t,a,c);
                update(k,t,b+1,-c);
            }
            else if(op==2)
            {
                int ans=0;
                scanf("%d",&a);
                for(int j=1;j<=10;j++) ans+=query(j,a%j,a);
                printf("%d\n",ans+num[a]);
            }
        }
    }
    return 0;
}