HDU 4267 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5339    Accepted Submission(s): 1693

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

 

 

 

解析：55个树状数组。

 

 

 

#include <cstdio>
#include <cstring>
#define lowbit(x) (x)&(-x)

const int MAXN = 50000+5;
int num[MAXN];
int c[MAXN][11][11];
int n;

void add(int x, int k, int mod, int val)
{
    for(int i = x; i <= n; i += lowbit(i))
        c[i][k][mod] += val;
}

int sum(int x, int a)
{
    int ret = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        for(int j = 1; j <= 10; ++j)
            ret += c[i][j][a%j];
    return ret;
}

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