A Simple Problem with Integers+hdu+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): 3924    Accepted Submission(s): 1213

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
解决方案：本题可用树状数组来做，可是更新线段的时候有个限定的条件a<=i<=b,且(i-a)%k==0，如果单点的话时间复杂度是很高的。所以我们可以做个多状态的树状数组C[i][k][mod]，表示除k，余mod，a%k==mod,更新的是(b,k,a%k,c)表示1到b全加c,(a-1,k,a%k,-c)表示1到(a-1)减c，状态时k、a%k。由于(i-a)%k==0查询的时候可以枚举1到10这几个k值，把所有相关状态加起来即可。
code：
#include <iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn=55000;int C[maxn][11][11],n;int num[maxn];void add(int x,int k,int mod,int v){    while(x>0)    {        C[x][k][mod]+=v;        x-=(x&(-x));    }}int query(int x,int a){    int ret=0;    while(x<=n)    {        for(int i=1; i<=10; i++)            ret+=C[x][i][a%i];        x+=(x&(-x));    }    return ret;}int main(){    while(~scanf("%d",&n))    {        int q,op,a,b,k,c;        memset(C,0,sizeof(C));        for(int i=1; i<=n; i++)        {            scanf("%d",&num[i]);        }        scanf("%d",&q);        for(int i=1; i<=q; i++)        {            scanf("%d",&op);            if(op==1)            {                scanf("%d%d%d%d",&a,&b,&k,&c);                add(b,k,a%k,c);                add(a-1,k,a%k,-c);            }            else            {                scanf("%d",&a);                printf("%d\n",query(a,a)+num[a]);            }        }    }    return 0;}