HDU-4267-A Simple Problem with Integers-(树状数组)
2017-10-05 16:54
573 查看
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
4000
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
树状数组很有趣的一种做法
代码:
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
4000
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
树状数组很有趣的一种做法
代码:
#include<iostream> #include<string> #include<cstdio> #include<algorithm> #include<cmath> #include<iomanip> #include<queue> #include<cstring> #include<map> using namespace std; typedef long long ll; #define M 50005 int tree[M][11][11]; int p[M]; int n; inline int lowbit(int i) { return i&(-i); } void update(int x,int k,int mod,int v) { while(x<=n) { tree[x][k][mod]+=v; x+=lowbit(x); } } int query(int x,int y) { int ret=0,i; while(x>0) { for(i=1;i<=10;i++) { ret+=tree[x][i][y%i]; } x-=lowbit(x); } return ret; } int main() { int q,i,op,a,b,k,c; while(scanf("%d",&n)!=EOF) { for(i=1;i<=n;i++) scanf("%d",&p[i]); memset(tree,0,sizeof(tree)); scanf("%d",&q); while(q--) { scanf("%d",&op); if(op==2) { scanf("%d",&a); int ans=query(a,a); printf("%d\n",p[a]+ans); } else { scanf("%d%d%d%d",&a,&b,&k,&c); update(a,k,a%k,c); update(b+1,k,a%k,-c); } } } return 0; }
相关文章推荐
- A Simple Problem with Integers 多树状数组分割,区间修改,单点求职。 hdu 4267
- 【三维树状数组与离散化】HDU 4267——A Simple Problem with Integers
- HDU 4267 A Simple Problem with Integers (树状数组)
- HDU 4267 A Simple Problem with Integers(树状数组)
- HDU 4267 A Simple Problem with Integers【树状数组】
- HDU 4267 - A Simple Problem with Integers 树状数组区间修改
- HDU - 4267 A Simple Problem with Integers(树状数组的逆操作)
- HDU 4267 A Simple Problem with Integers(树状数组)
- HDU - 4267 A Simple Problem with Integers(树状数组)
- hdu 4267 A Simple Problem with Integers(分类别维护多个树状数组)
- HDU 4267 A Simple Problem with Integers [树状数组]
- HDU 4267 A Simple Problem with Integers 多个树状数组
- A Simple Problem with Integers+hdu+树状数组
- HDOJ 4267 A Simple Problem with Integers 树状数组
- HDU 4267 - A Simple Problem with Integers
- HDU 4267-A Simple Problem with Integers(多个BIT)
- POJ 3468 A Simple Problem with Integers 树状数组 区间修改 区间查询
- 模板(线段树 + 树状数组 + 区间修改 + 区间查询)eg:POJ 3468 - A Simple Problem with Integers
- HDU 4267 A Simple Problem with Integers --树状数组
- poj 3486 A Simple Problem with Integers(树状数组第三种模板改段求段)