HDU-4267-A Simple Problem with Integers-（树状数组）

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 1142 12 22 32 41 2 3 1 22 1 2 22 32 41 1 4 2 12 12 22 32 4

 

Sample Output

111113312341

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 50005int 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;}