Hduoj【树状数组】【-.-】

/*A Simple Problem with IntegersTime Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3885    Accepted Submission(s): 1197Problem DescriptionLet 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. InputThere 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)OutputFor each test case, output several lines to answer all query operations.Sample Input4 1 1 1 1142 12 22 32 41 2 3 1 22 1 2 22 32 41 1 4 2 12 12 22 32 4 Sample Output111113312341Source2012 ACM/ICPC Asia Regional Changchun Online */#include<stdio.h>#include<string.h>int a1[50010], c1[11][11][50010];int lowbit(int x){return x&(-x);}int sum(int x, int a, int n){int sum = 0, i;while(x <= n){for(i  = 1; i <= 10 ; i++){sum += c1[i][a%i][x];}x += lowbit(x);}return sum;}void change(int i, int k, int k1, int e){while(i > 0){c1[k][k1][i] += e;i -= lowbit(i);}}int main(){int i, j, k, m, n;while(scanf("%d", &n) != EOF){memset(c1, 0, sizeof(c1));for(i = 1; i <= n; i++)scanf("%d", &a1[i]);scanf("%d", &m);while(m--){int o, a, b, c, d;scanf("%d", &o);if(o == 1){scanf("%d%d%d%d", &a, &b , &c, &d);change(a-1, c, a%c, -d);//a前面的全-d change(b, c, a%c, d);//b到前面的全+d }else{scanf("%d", &a);printf("%d\n", sum(a,a,n) + a1[a]);}} }return 0;}

题意：给出一个数组，分别对每隔k个点进行加减，并且要实现点查询。

体会 ：难，我只是套用了别人的模板，没理解到底什么意思，等来年有机会了定回头来搞定它。