A Simple Problem with Integers(100棵树状数组)
A Simple Problem with Integers
Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5034 Accepted Submission(s): 1589
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,每次更新(i - a)%k = 0 的位置的值,每次询问a位置的值;所以i%k = a%k;
开100棵树状数组,tree[x][k][mod];
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<cmath> using namespace std; const int MAXN = 50100; int tree[MAXN][11][11]; int num[MAXN]; int lowbit(int x){return x & (-x);} void add(int x, int k, int mod, int v){ while(x < MAXN){ tree[x][k][mod] += v; x += lowbit(x); } } int sum(int x, int a){ int ans = 0; while(x > 0){ for(int i = 1; i <= 10; i++){ ans += tree[x][i][a % i]; } x -= lowbit(x); } return ans; } int main(){ int N, M; while(~scanf("%d", &N)){ memset(tree, 0, sizeof(tree)); for(int i = 1; i <= N; i++){ scanf("%d", &num[i]); } scanf("%d", &M); int q, a, b, k, c; while(M--){ scanf("%d%d", &q, &a); if(q == 1){ scanf("%d%d%d", &b, &k, &c); add(a, k, a % k, c); add(b + 1, k, a % k, -c); } else{ printf("%d\n", num[a] + sum(a, a)); } } } return 0; }