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][12][12];

int a[maxn];

int n;

int lowbit(int x)

{

    return x&(-x);

}

void update2(int x,int num,int k,int mod)

{

    while(x>0)

    {

        d[x][k][mod]+=num;

        x-=lowbit(x);

    }

}

int getSum1(int x,int k)

    int s=0;

    while(x<=n)

    {

        REP_1(i,10)

        {

            s+=d[x][i][k%i];

        }

        x+=lowbit(x);

    }

    return s;

}



int main()

{

    while(RD(n)!=-1)

    {

        REP_1(i,n)

            RD(a[i]);

        memset(d,0,sizeof(d));

        int q;

        RD(q);

        while(q--)

        {

            int t;

            RD(t);

            if(t==1)

            {

                int l,r,k,c;

                RD(l),RD(r),RD(k),RD(c);

                update2(r,c,k,l%k);

                update2(l-1,-c,k,l%k);

            }

            if(t==2)

            {

                int c;

                RD(c);

                printf("%d\n",getSum1(c,c)+a[c]);

            }

        }

    }

}