HDU 4267

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 323 Accepted Submission(s): 133

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

 1 //原来我用插点问线做的，输出sum(temp) - sum(temp-1),一直tle 

 2 #include <stdio.h>

 3 #include <iostream>

 4 #include<string.h>

 5 using namespace std;

 6 

 7 const int MAXD = 50010;

 8 int N, M, d[100][MAXD], a[MAXD];

 9 

10 void insert(int k, int x, int v)

11 {

12      for(; x <= N; x += x & -x) 

13           d[k][x] += v;

14 }

15 

16 void init()

17 {

18     int i, j;

19      for(i = 1; i <= N; i ++)

20           cin>>a[i];

21      for(i = 0; i < 100; i ++) 

22           memset(d[i], 0, sizeof(d[i][0]) * (N + 1));

23 }

24 

25 int query(int id)

26 {

27     int i, k, x, ans = 0;

28     for(i = 1; i <= 10; i ++)

29     {

30         k = (i - 1) * 10 + id % i;

31         for(x = id; x > 0; x -= x & -x) 

32           ans += d[k][x];

33     }

34     return a[id] + ans;

35 }

36 

37 void solve()

38 {

39     int i, flag,temp;

40     int a, b, k, c;

41     cin>>M;

42     while(M--)

43     {

44         cin>>flag;

45         if(flag == 1)

46         {

47             cin>>a>>b>>k>>c;

48             b -= (b - a) % k;

49             insert(10 * (k - 1) + a % k, a, c);

50             if(b < N) insert(10 * (k - 1) + b % k, b + 1, -c);

51         }

52         else if(flag == 2)

53         {

54             cin>>temp;

55             cout<<query(temp)<<endl;

56         }

57     }

58 }

59 

60 int main()

61 {

62     while(cin>>N)

63     {

64         init();

65         solve();

66     }

67     return 0;

68 }