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

 

 

 

解析：55个树状数组。

 

 

 

#include 
      
       #include 
       
        #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;}