A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6241 Accepted Submission(s): 2012

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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题意：

给了N个数，有一些操作。

操作一是让a到b区间的满足(i-a)%k=0的数，加上c。

操作二是询问位置为I的数为多少。

思路：

借鉴这个http://blog.sina.com.cn/s/blog_a047b3eb010171cc.html

想到树状数组有一种技巧，我们可以对a和b+1两个点单点更新，让a位置+c，b+1位置-c，这样在用树状数组求和的时候，对于a到b的点都是加上了c。所以这样我们让a到b区间无限制的都加上了c，但是我们只想让满足(i-a)%k=0的位置上的点+c。

由(i-a)%k=0可以推出 i%k=a%k，也就是只有当i%k恰好等于a%k时这些数才加上了c。所以我们用树状数组查询的时候，只查满足a%k的数。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=50005;
typedef long long ll;
int tr[maxn][11][10];
int arr[maxn];
int lowbit(int i)
{
    return i&(-i);
}
int query(int n)
{
    int sum=0;
    for(int j=1;j<=10;j++)
    {
        for(int i=n;i>0;i-=lowbit(i))
        {
            sum+=tr[i][j][n%j];
        }
    }
    return sum+arr[n];
}
void update(int a,int k,int c,int mod,int n)
{
    for(int i=a;i<=n;i+=lowbit(i))
    {
        tr[i][k][mod]+=c;
    }
}
int main()
{
    int n,q,a,b,k,c,op;
    while(~scanf("%d",&n))
    {
        memset(tr,0,sizeof(tr));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&arr[i]);
        }
        scanf("%d",&q);
        for(int i=1;i<=q;i++)
        {
            scanf("%d",&op);
            if(op==2)
            {
                scanf("%d",&a);
                printf("%d\n",query(a));
            }
            else
            {
                scanf("%d%d%d%d",&a,&b,&k,&c);
                update(a,k,c,a%k,n);
                update(b+1,k,-c,a%k,n);
            }
        }
    }
    return 0;
}

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hdu 4267 A Simple Problem with Integers(树状数组)

A Simple Problem with Integers

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