HDU A Simple Problem with Integers

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：
      
      
      
      
首先给n个数字A1，A2...An，然后会进行两种操作
      
      
      
      
操作1：a b k c表示a<=i<=b中（i-a）%k == 0就加c。
      
      
      
      
操作2：a 查询A[a]的值。
      
      
      
      
 
      
      
      
      
此题我用线段写了一遍，本以为维护左边界和取余值就可以了，但是一直错（水平不够吧），
      
      
      
      
然后看了一下题解，因为k<=10，所以取余k的余数种数最多55种，所以可以用一个二维
      
      
      
      
数组用树状数组维护，将a%k映射为（k-1）*10+a%k，这样保证不同k的不同余数的下标值唯一，
      
      
      
      
然后将a更新c，将b+1更新为-c，这样保证求前缀和的时候如果在[a,b]内就会加c，不在就不会加，
      
      
      
      
然后是为什么可以直接查询在[a,b]间就加呢？因为如果a%k == b%k，那么（b-a）%k == 0，
      
      
      
      
映射a的时候是(i-1)*10+a%i就保证余数相等了，就只需要考虑在不在更新的区间内就行了。
      
      
      
      

      
      
      
      
#include <stdio.h>
#include <string.h>
const int maxn = 50005;
int sum[100][maxn], A[maxn], n;
inline int lowbit ( int x ) { return x & -x; }
void add ( int k, int x, int v )
{
    while ( x <= n )
    {
        sum[k][x] += v;
        x += lowbit ( x );
    }
}
int bit_sum ( int x )
{
    int res = 0;
    for ( int i = 1; i <= 10; i ++ )
    {
        //a%k == b%k  -->  ( b-a )%k == 0
        //所以只需要知道在区间更新的值为多少
        int k = ( i-1 )*10+x%i; //所有起始位置余数为x%i的和
        for ( int j = x; j > 0; j -= lowbit ( j ) )
            res += sum[k][j];
    }
    return res+A[x];
}
void solve ( )
{
    while ( ~ scanf ( "%d", &n ) )
    {
        memset ( sum, 0, sizeof ( sum ) );
        for ( int i = 1; i <= n; i ++ )
            scanf ( "%d", &A[i] );
        int q, op, a, b, k, c;
        scanf ( "%d", &q );
        while ( q -- )
        {
            scanf ( "%d", &op );
            if ( op == 1 )
            {
                scanf ( "%d%d%d%d", &a, &b, &k, &c );
                add ( 10*( k-1 )+a%k, a, c );   //将余数映射到下标
                add ( 10*( k-1 )+a%k, b+1, -c );
            }
            else
            {
                scanf ( "%d", &a );
                printf ( "%d\n", bit_sum ( a ) );
            }
        }
    }
}
int main ( )
{
    solve ( );
    return 0;
}