HDU - 4267 A Simple Problem with Integers（树状数组的逆操作）

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：给你两个操作1 a b k c代表的是将区间[a, b]之间的能使得(i-a)%k = 0的数加上c，2 a是查询值

思路：用树状数组做，但是和一般的树状数组不一样，这里是将sum和add操作交换了（这点需要好好理解一下，我们通过两次的以前的sum操作将区间[a, b]赋值，然后再通过以前的add操作找到可以影响到它的结点求总的操作值），将取模公式转换成：i%k=a%k,那么我们注意到k是比较小的，所以我们可以先记录a%k时操作的结果，也就是开arr[x][k][a%k]表示，最后再求和计算x坐标的结果是枚举余数就行了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 50005;

int arr[maxn][11][11];
int n, num[maxn];

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

inline void Add(int x, int k, int mod, int c){
	while (x > 0) {
		arr[x][k][mod] += c;
		x -= lowbit(x);
	}
}  

inline int sum(int x, int a) {
	int ans = 0;
	while (x < maxn) {
		for (int i = 1; i <= 10; i++){
			ans += arr[x][i][a%i];
		}
		x += lowbit(x);
	}
	return ans;
}  

int main() {
	while (scanf("%d", &n) != EOF) {
		for (int i = 1; i <= n; i++)
			scanf("%d", &num[i]);
		memset(arr, 0, sizeof(arr));
		int m;
		scanf("%d", &m);
		while (m--) {
			int op, a, b, k, c;
			scanf("%d", &op);
			if (op == 1) {
				scanf("%d%d%d%d", &a, &b, &k, &c);
				Add(b, k, a%k, c);
				Add(a-1, k, a%k, -c);
			}
			else{
				scanf("%d", &a);
				int ans = sum(a, a);
				printf("%d\n", ans + num[a]);
			}
		}
	}
	return 0;
}