HDU 4267 A Simple Problem with Integers

 

                      A Simple Problem with Integers

                                              Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
                                                                   Total Submission(s): 2037 Accepted Submission(s): 673

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

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liuyiding

      
这题刚开始看到的时候，以为是用线段树做，实际上用线段树完全可以解出来，但是没有想法，就想是不是方法错了，后来发现或许最大流可以，但是用最大流肯定会超时，只是建图就超时了，就更别说其他的了。然后实在没有办法的情况下，搜了一下解题报告，看到能用树状数组做，结果自己不会这算法，就重新开始学这算法，学好了后，看解题报告的时候，看到人家的代码和那算法的代码几乎就是一样的，少部分做了修改。可他们的思路我实在是想不通。后来终于找到了一个我能理解的了。 他的update() 函数，和求和函数，里面的循环次序与   树状数组正好倒了过来。举了个例子正好对。神一样的思路啊。

#include <stdio.h>
#include <math.h>
#include <string.h>
int tree[12][12][50010];
int a[50010],n;
int main()
{
    void update(int k,int mod,int border,int val);
    int sum(int k,int mod,int border);
    int i,j,m,s,t,l,r,k,val;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(tree,0,sizeof(tree));
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d",&t);
            if(t==1)
            {
                scanf("%d %d %d %d",&l,&r,&k,&val);
                update(k,l%k,r,val);
                update(k,l%k,l-1,-val);
            }else
            {
                scanf("%d",&l);
                s=a[l];
                for(i=1;i<=10;i++)
                {
                    s+=sum(i,l%i,l);
                }
                printf("%d\n",s);
            }
        }
    }
    return 0;
}
int lowbit(int x)
{
    return (x&-x);
}
void update(int k,int mod,int border,int val)
{
    while(border>0)
    {
        tree[k][mod][border]+=val;
        border=border- lowbit(border);
    }
}
int sum(int k,int mod,int border)
{
    int s=0;
    while(border<=n)
    {
        s+=tree[k][mod][border];
        border=border + lowbit(border);
    }
    return s;
}