HDOJ 4267 A Simple Problem with Integers 树状数组

建10*10个树状数组, tree[x][y] 表示从x开始等差为y的树状数组.对于每个询问a,统计与a相关的的数组数组的和.

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A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4566 Accepted Submission(s): 1406

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

/* ***********************************************
Author        :CKboss
Created Time  :2015年09月07日 星期一 22时43分29秒
File Name     :HDOJ4267.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

const int maxn=50500;

inline int lowbit(int x) { return x&(-x); }

int color[12][12][maxn];
int tree[12][12][maxn];
int C=0;

void add(int x,int y,int p,int v)
{
    for(int i=p;iif(color[x][y][i]!=C)
        {
            tree[x][y][i]=0;
            color[x][y][i]=C;
        }
        tree[x][y][i]+=v;
    }
}

int sum(int x,int y,int p)
{
    int ret=0;
    for(int i=p;i>0;i-=lowbit(i))
    {
        if(color[x][y][i]==C)
            ret+=tree[x][y][i];
    }
    return ret;
}

int calu(int u)
{
    int ret=0;
    for(int x=1;x<=10;x++)
    {
        for(int y=1;y<=10;y++)
        { 
            if((u-x)%y==0)
            {
                ret+=sum(x,y,u);
            }
        }
    }
    return ret;
}

int n,m,A[maxn];
int kind,a,b,k,c;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++) scanf("%d",A+i);
        C++;
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d",&kind);
            if(kind==1)
            {
                scanf("%d%d%d%d",&a,&b,&k,&c);
                int x=(a%k); if(x==0) x=k;
                add(x,k,a,c);
                add(x,k,b+1,-c);
            }
            else if(kind==2)
            {
                scanf("%d",&a);
                printf("%d\n",calu(a)+A[a]);
            }
        }
    }

    return 0;
}