hdu 5057 Argestes and Sequence(分块算法)

Argestes and Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 566    Accepted Submission(s): 142

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]<=$2^{31}$ - 1
1<=X<=N
0<=Y<=$2^{31}$ - 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

 

分块算法：大致就是把一堆(包含n个基本单位)东西分成sqrt(n)块， 每块包含sqrt(n)个基本单位。

如果要修改某个基本单位，只要修改该单位所在的块，因为每块包含sqrt(n)个基本单位，所以时间复杂度为sqrt(n)；

view code#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 100010;

struct Block
{
    int ct[10][10];
}block[400];
int _, n, m, a[N], cnt, pp[11];

void build()
{
    int tmp = (int)sqrt(n*1.0), id;
    cnt = n/tmp + 1;

    memset(block, 0, sizeof(block));
    for(int i=0; i<n; i++)
    {
        scanf("%d", &a[i]);
        id = i/cnt, tmp = a[i];
        for(int j=0; j<10; j++)
        {
            block[id].ct[j][tmp%10]++;
            tmp /= 10;
        }
    }
}

void update(int u, int v)
{
    int id = u/cnt;
    for(int i=0; i<10; i++)
    {
        block[id].ct[i][a[u]%10]--;
        a[u] /= 10;
    }
    a[u] = v;
    for(int i=0; i<10; i++)
    {
        block[id].ct[i][v%10]++;
        v /= 10;
    }
}

int query(int l, int r, int d, int p)
{
    int L = l/cnt, R = r/cnt;
    int res = 0, div = pp[d];
    if(L==R)
    {
        for(int i=l; i<=r; i++) if(a[i]/div%10==p) res++;
        return res;
    }

    for(int i=L+1; i<R; i++)
    {
        res += block[i].ct[d][p];
    }

    for(int i=l; i<(L+1)*cnt; i++)
    {
        if(a[i]/div%10==p) res++;
    }

    for(int i=R*cnt; i<=r; i++)
    {
        if(a[i]/div%10==p) res++;
    }
    return res;
}

void solve()
{
    scanf("%d%d", &n, &m);
    build();

    char str[5];
    int u, v, d, p;
    while(m--)
    {
        scanf("%s", str);
        if(str[0]==‘S‘)
        {
            scanf("%d%d", &u, &v);
            u--;
            update(u, v);
        }
        else
        {
            scanf("%d%d%d%d", &u, &v, &d, &p);
            u--, v--, d--;
            printf("%d\n", query(u, v, d, p));
        }
    }
}

void init()
{
    pp[0] = 1;
    for(int i=1; i<10; i++) pp[i] = pp[i-1]*10;
}

int main()
{
//    freopen("in.txt", "r", stdin);
    init();
    cin>>_;
    while(_--) solve();
    return 0;
}