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 = 100005;
int a[N], len = 400, n, mi[15] = {0, 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000};

struct node
{
    int cnt[12][12];//cnt[d][p]表示每个块内第d位是p的个数
}block[400];

bool judge(int x, int d, int p)//判断x的第d位是否为p
{
    return x/mi[d]%10==p;
}

int query(int l, int r, int d, int p)
{
    int id1 = l/len, id2 = r/len, ans = 0;
    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+1; i <= id2-1; i++)//把中间的块加起来
          ans+=block[i].cnt[d][p];
        for(int i = l; i <= (id1+1)*len && i <= n; i++)//暴力最左边的块
          if(judge(a[i], d, p))
            ans++;
        for(int i = id2*len+1; 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-1)/len;//这里起初x少减了个1，WA了好多次忧伤。。。
    a[x] = y;
    for(int i = 1; i <= 10; i++)
    {
        block[id].cnt[i][tmp%10]--;
        tmp/=10;
    }
    tmp = a[x];
    for(int i = 1; i <= 10; i++)
    {
        block[id].cnt[i][tmp%10]++;
        tmp/=10;
    }
}

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, 0, sizeof(block));
        memset(a, 0, sizeof(a));
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            id = (i-1)/len;
            int tmp = a[i];
            for(int j = 1; j <= 10; j++)//初始化块
            {
                block[id].cnt[j][tmp%10]++;
                tmp/=10;
            }
        }
        while(m--)
        {
            getchar();
            scanf("%c", &cmd);
            if(cmd == 'Q')
            {
                int ans = 0;
                scanf("%d%d%d%d", &l, &r, &d, &p);
                ans = query(l, r, d, p);
                printf("%d
", ans);
            }else
            {
                scanf("%d%d", &x, &y);
                update(x, y);
            }
        }
    }

    return 0;
}