HDU4267 树状数组 不连续区间修改(三维)

A Simple Problem with Integers

                                

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

题解：

     树状数组的区间修改，单点查询，注意这里的不连续区间修改

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
const int  N =  + ;

int C[][][N],n;
int getsum(int x,int a) {
    int sum = ;
    while(x > ) {
       for(int i = ; i <= ; i++) sum += C[i][a%i][x];
        x -= x & (-x);
    }
    return sum;
}
void update(int a,int b,int x,int v) {
    while(x <= n) {
        C[a][b][x] += v;
        x += x & (-x);
    }
}
int main() {
    int o[N],a,b,c,k,q,t;
    while(~scanf("%d", &n)) {
            memset(C,,sizeof(C));
        for(int i = ; i <= n; i++) scanf("%d", & o[i]);
        scanf("%d", &q);
        while(q --) {
            scanf("%d", &t);
            if(t == ) {
                scanf("%d%d%d%d",&a,&b,&k,&c);
                update(k,a%k,a,c);
                update(k,a%k,b+,-c);
            }
            else {
                scanf("%d", &a);
                printf("%d\n", getsum(a,a) + o[a]);
            }
        }
    }
    return ;
}