A Simple Problem with Integers 多树状数组分割，区间修改，单点求职。 hdu 4267

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4032    Accepted Submission(s): 1255

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 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

 

(i-a)%k==0   即  i%k==a%k             分组   x%k==a%k的为一组，   参数 mod, k,x

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <iomanip>
#include <sstream>
using namespace std;
//#define local
typedef long long LL;
const int INF=0x4fffffff;
const int EXP=1e-5;
const int MS=50005;

int C[11][11][MS];      //C[mod][k][x]
int num[MS];

int lowbit(int x)
{
      return x&(-x);
}

//   修改区间，单点求职，   树状数组需要逆过来。

void updata(int mod,int k,int x,int d)
{
      while(x>0)
      {
            C[mod][k][x]+=d;
            x-=lowbit(x);
      }
}

int getsum(int a,int x)
{
      int res=0;
      while(x<MS)     //x<=n
      {
            for(int k=1;k<=10;k++)
            {
                  res+=C[a%k][k][x];
            }
            x+=lowbit(x);
      }
      return res;
}

int main()
{
      #ifdef local
      freopen("in.txt","r",stdin);
      freopen("out.txt","w",stdout);
      #endif // local
      int n;
      while(scanf("%d",&n)!=EOF)
      {
            for(int i=1;i<=n;i++)
                  scanf("%d",&num[i]);
            memset(C,0,sizeof(C));
            int m;
            scanf("%d",&m);
            while(m--)
            {
                  int op,a,b,k,c;
                  scanf("%d",&op);
                  if(op==1)
                  {
                        scanf("%d%d%d%d",&a,&b,&k,&c);
                        updata(a%k,k,b,c);
                        updata(a%k,k,a-1,-c);
                  }
                  else
                  {
                        scanf("%d",&a);
                        int ans=getsum(a,a);
                        printf("%d\n",ans+num[a]);
                  }
            }
      }
      return 0;
}