hdu4267A Simple Problem with Integers【树状数组区间更新/单点查询】

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
   
   
   
   

 

终于把区间更新搞明白了，书上的写法实在是晦涩难懂，明明是区间更新，干嘛非要生搬硬套单点更新的写法？？区间更新，带入函数的参数是向下管理从1到这个点的所有区域的，而单点更新，已知了一个点，管理这个点的高层们都需要更新自己的值，这就是本质区别，而且选择适合的算法时间也要稍微快一点

/*************
hdu4267
2016.2.1
483MS	29952K	1447 B	G++old
374MS	29796K	1390 B	G++new
*************/
#include<cstdio>
#include<cstring>
using namespace std;
int n,q,c[12][12][50010],tmp[50010];
int lowbit(int i)
{
    return i&(-i);
}
void add(int t1,int t2,int i,int x)
{
    while(i>0)
    {
        c[t1][t2][i]+=x;
        i-=lowbit(i);
    }
}
int query(int t1,int t2,int x)
{
    int s=0;
    while(x<=n)
    {
        s+=c[t1][t2][x];
        x+=lowbit(x);
    }
    return s;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        int a,b,k,e,m;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&tmp[i]);
        }
   //     for(int i=1;i<=n;i++) printf("%d  ",c[i]);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d",&m);
            if(m==1)
            {
                scanf("%d%d%d%d",&a,&b,&k,&e);
                int num=(b-a)/k;
                int s=a%k;
                add(k,s,a-1,-e);
                add(k,s,b,e);
            }
            else
            {
                scanf("%d",&a);
                int sum=tmp[a];
                for(int i=1;i<=10;i++)
                {
                    sum+=query(i,a%i,a);
                }
                printf("%d\n",sum);
            }
        }
    }
    return 0;
}

可恶的点更新的算法

/*************
hdu4267
2016.2.1
*************/
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,q,c[12][12][50010],tmp[50010];
int lowbit(int i)
{
    return i&(-i);
}
void add(int t1,int t2,int i,int x)
{
    while(i<=n)
    {
        c[t1][t2][i]+=x;
        i+=lowbit(i);
    }
}
int query(int t1,int t2,int x)
{
    int s=0;
    while(x>0)
    {
        s+=c[t1][t2][x];
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        int a,b,k,e,m;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&tmp[i]);
        }
   //     for(int i=1;i<=n;i++) printf("%d  ",c[i]);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d",&m);
            if(m==1)
            {
                scanf("%d%d%d%d",&a,&b,&k,&e);
                a--;b--;
                int num=(b-a)/k;
                int s=a%k;
                add(k,s,a/k+1,e);
                add(k,s,a/k+num+2,-e);
            }
            else
            {
                scanf("%d",&a);
                a--;
                int sum=tmp[a];
                for(int i=1;i<=10;i++)
                {
                    sum+=query(i,a%i,a/i+1);
                }
                printf("%d\n",sum);
            }
        }
    }
    return 0;
}