HDU5057(分块)

　　Argestes and Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1279    Accepted Submission(s): 373

Problem Description

Argestes has a lot of hobbies and likes solving query problems especially. One day Argestes came up with such a problem. You are given a sequence a consisting of N nonnegative integers, a[1],a[2],...,a[n].Then there are M operation on the sequence.An operation can be one of the following:
S X Y: you should set the value of a[x] to y(in other words perform an assignment a[x]=y).
Q L R D P: among [L, R], L and R are the index of the sequence, how many numbers that the Dth digit of the numbers is P.
Note: The 1st digit of a number is the least significant digit.

 

Input

In the first line there is an integer T , indicates the number of test cases.
For each case, the first line contains two numbers N and M.The second line contains N integers, separated by space: a[1],a[2],...,a[n]—initial value of array elements.
Each of the next M lines begins with a character type.
If type==S,there will be two integers more in the line: X,Y.
If type==Q,there will be four integers more in the line: L R D P.

[Technical Specification]
1<=T<= 50
1<=N, M<=100000
0<=a[i]<=231 - 1
1<=X<=N
0<=Y<=231 - 1
1<=L<=R<=N
1<=D<=10
0<=P<=9

 

Output

For each operation Q, output a line contains the answer.

 

Sample Input

1

5 7

10 11 12 13 14

Q 1 5 2 1

Q 1 5 1 0

Q 1 5 1 1

Q 1 5 3 0

Q 1 5 3 1

S 1 100

Q 1 5 3 1

Sample Output

5

1

1

5

0

1

 

分块大法，降低暴力的时间复杂度

 //2016.8.13
 #include<iostream>
 #include<cstdio>
 #include<cstring>

 using namespace std;

 const int N = ;
 int a[N], len = , n, mi[] = {, , , , , , , , , , };

 struct node
 {
     int cnt[][];//cnt[d][p]表示每个块内第d位是p的个数
 }block[];

 bool judge(int x, int d, int p)//判断x的第d位是否为p
 {
     return x/mi[d]%==p;
 }

 int query(int l, int r, int d, int p)
 {
     int id1 = l/len, id2 = r/len, ans = ;
     if(id1==id2)//如果l,r在同一个块内，暴力枚举
     {
         for(int i = l; i <= r; i++)
               if(judge(a[i], d, p))
                   ans++;
     }else
     {
         for(int i = id1+; i <= id2-; i++)//把中间的块加起来
           ans+=block[i].cnt[d][p];
         for(int i = l; i <= (id1+)*len && i <= n; i++)//暴力最左边的块
           if(judge(a[i], d, p))
             ans++;
         for(int i = id2*len+; i <= r; i++)//暴力最右边的块
           if(judge(a[i], d, p))
             ans++;
     }
     return ans;
 }

 void update(int x, int y)
 {
     int tmp = a[x], id = (x-)/len;//这里起初x少减了个1，WA了好多次忧伤。。。
     a[x] = y;
     for(int i = ; i <= ; i++)
     {
         block[id].cnt[i][tmp%]--;
         tmp/=;
     }
     tmp = a[x];
     for(int i = ; i <= ; i++)
     {
         block[id].cnt[i][tmp%]++;
         tmp/=;
     }
 }

 int main()
 {
     int T, d, p, m, x, y, l, r, id;
     char cmd;
     cin>>T;
     while(T--)
     {
         scanf("%d%d", &n, &m);
         memset(block, , sizeof(block));
         memset(a, , sizeof(a));
         for(int i = ; i <= n; i++)
         {
             scanf("%d", &a[i]);
             id = (i-)/len;
             int tmp = a[i];
             for(int j = ; j <= ; j++)//初始化块
             {
                 block[id].cnt[j][tmp%]++;
                 tmp/=;
             }
         }
         while(m--)
         {
             getchar();
             scanf("%c", &cmd);
             if(cmd == 'Q')
             {
                 int ans = ;
                 scanf("%d%d%d%d", &l, &r, &d, &p);
                 ans = query(l, r, d, p);
                 printf("%d\n", ans);
             }else
             {
                 scanf("%d%d", &x, &y);
                 update(x, y);
             }
         }
     }

     return ;
 }