HDU 4267 A Simple Problem with Integers 多个树状数组

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4000    Accepted Submission(s): 1243

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

 

 

题意

就是给你n个数，m个操作

操作分为两种

1 a b k c,就是 i属于(a,b)这个区间，而且i必须满足(i-a)%k==0，然后这个数加上c

2 a，问你坐标为a的数的大小是多少

 

 

题解

他是分段求和，分段加，肿么办！

那我们建立Ｎ多树状数组就好啦，然后直接区间加加加加！！！

然后就好了

 

代码

int d[maxn][][];
int a[maxn];
int n;
int lowbit(int x)
{
    return x&(-x);
}
void update2(int x,int num,int k,int mod)
{
    while(x>)
    {
        d[x][k][mod]+=num;
        x-=lowbit(x);
    }
}
int getSum1(int x,int k)
    int s=;
    while(x<=n)
    {
        REP_1(i,)
        {
            s+=d[x][i][k%i];
        }
        x+=lowbit(x);
    }
    return s;
}

int main()
{
    while(RD(n)!=-)
    {
        REP_1(i,n)
            RD(a[i]);
        memset(d,,sizeof(d));
        int q;
        RD(q);
        while(q--)
        {
            int t;
            RD(t);
            if(t==)
            {
                int l,r,k,c;
                RD(l),RD(r),RD(k),RD(c);
                update2(r,c,k,l%k);
                update2(l-,-c,k,l%k);
            }
            if(t==)
            {
                int c;
                RD(c);
                printf("%d\n",getSum1(c,c)+a[c]);
            }
        }
    }
}