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-;
 const int MS=;

 int C[][][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>)
       {
             C[mod][k][x]+=d;
             x-=lowbit(x);
       }
 }

 int getsum(int a,int x)
 {
       int res=;
       while(x<MS)     //x<=n
       {
             for(int k=;k<=;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=;i<=n;i++)
                   scanf("%d",&num[i]);
             memset(C,,sizeof(C));
             int m;
             scanf("%d",&m);
             while(m--)
             {
                   int op,a,b,k,c;
                   scanf("%d",&op);
                   if(op==)
                   {
                         scanf("%d%d%d%d",&a,&b,&k,&c);
                         updata(a%k,k,b,c);
                         updata(a%k,k,a-,-c);
                   }
                   else
                   {
                         scanf("%d",&a);
                         int ans=getsum(a,a);
                         printf("%d\n",ans+num[a]);
                   }
             }
       }
       return ;
 }