hdu4267 A Simple Problem with Integers

题目大意：解释起来有些麻烦，直接贴一下英文版好了

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)

 

如果k只能等于1，那么这就是一道很简单的题，对于每个操作1 a b 1 c，只需要将p[a] += c，p[b+1] -= c。对于查询2 x，只要统计p[1]+p[2]+p[3]+.......p[x-1]+p[x]的值就好了。

由此出发，可以发现这一题的k并不大，仅仅只有10，这样就可以对每个k做一种维护

 

具体做法是：

对于每个操作1 a b k c

h[k][a] += c, h[k][u] -= c , (u-a)%k == 0 且 u > b 且 u-k < b

对于每个查询2 x

就需要统计

h[1][x] + h[1][x-1] + ..... + h[1][1] +

h[2][x] + h[2][x-2] + ...... + h[2][x%2+2] + h[2][x%2] +

.......

h[10][x] + h[10][x-10] + ....... + h[10][x%10+10] + h[10][x%10]

+a[x]

这个可以使用树状数组来维护，只是在树状数组维护的过程中需要进行一个下标的转换：

我的树状数组维护的代码：

int trans(int x, int k){x = x / k + 1;return x;}int _trans(int x, int k, int p){x = (x - 1) * k + p;return x;}int lowbit(int x){return x&(-x);}int add1(int x, int k){int p;p = x % k;x = trans(x, k);x += lowbit(x);x = _trans(x, k, p);return x;}int sub1(int x, int k){int p;p = x % k;x = trans(x, k);x -= lowbit(x);if (x == 0)return 0;x = _trans(x, k, p);return x;}int query_k(int x, int k){int ans = 0;while (x > 0){ans += h[k][x];x = sub1(x, k);}return ans;}void updata(int x, int k, int val){while (x <= n){h[k][x] += val;x = add1(x, k);}}

 

最后附上这一题全部的代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>using namespace std;const int MAXM = 15;const int MAXN = 6e4;int h[MAXM][MAXN], a[MAXN];int n, m;int trans(int x, int k){x = x / k + 1;return x;}int _trans(int x, int k, int p){x = (x - 1) * k + p;return x;}int lowbit(int x){return x&(-x);}int add1(int x, int k){int p;p = x % k;x = trans(x, k);x += lowbit(x);x = _trans(x, k, p);return x;}int sub1(int x, int k){int p;p = x % k;x = trans(x, k);x -= lowbit(x);if (x == 0)return 0;x = _trans(x, k, p);return x;}int query_k(int x, int k){int ans = 0;while (x > 0){ans += h[k][x];x = sub1(x, k);}return ans;}void updata(int x, int k, int val){while (x <= n){h[k][x] += val;x = add1(x, k);}}int query(int x){int i, ans = 0;for (i=1; i<=10; i++)ans += query_k(x, i);return ans;}int main(){int i, j, x, y, k, c, tt, ans, ll;while (scanf("%d",&n)!=EOF){for (i=0; i<=n; i++)for (j=1; j<=10; j++)h[j][i] = 0;for (i=1; i<=n; i++)scanf("%d",&a[i]);scanf("%d",&m);for (i=0; i<m; i++){scanf("%d",&tt);if (tt == 1){scanf("%d%d%d%d",&x,&y,&k,&c);updata(x, k, c);ll = x + (y - x) / k * k;while (ll <= y) ll += k;updata(ll, k, -1*c);}else if (tt == 2){scanf("%d",&x);ans = query(x) + a[x];printf("%d\n",ans);}}}return 0;}